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0)
HARVARD COLLEGE
LIBRARY
GIFT OF THE
GRADUATE SCHOOL
OF EDUCATION
3 2044 096 988 944
SECONDARY-SCHOOL
MATHEMATICS
BY
ROBERT L. SHORT
HEAD OF THE DEPARTMENT OF MATHEMATICS
CLEVELAND TECHNICAL HIGH SCHOOL
AND
WILLIAM H. ELSON
SUPERINTENDENT OF SCHOOLS, CLEVELAND, OHIO
BOOK I
D. C. HEATH & CO., PUBLISHERS
BOSTON NEW YORK CHICAGO
ft * '> I i % f
M1V*«n COLL Flic UMMY
> - F SCI, JUL U -J ii-i. r;
Copyright, 1910,
By D. C. Heath & Co.
IDS
PREFACE
This text differs widely from that marked out by custom
and tradition. It treats the various branches of mathematics
more with reference to their unities and less as isolated enti-
ties (sciences). It seeks to give pupils usable knowledge of
the principles underlying mathematics and ready control of
them. These texts are not an experiment; they were thor-
oughly tried out in mimeograph form on hundreds of high
school pupils before being put into book form.
The scope of Books I and II does not vary greatly from •
that covered in algebras and geometries of the usual type.
However, Book I is different in that arithmetic, algebra, and
geometry are treated side by side. The effect of this arrange-
ment is increased interest and power of analysis on the part
of the learner, and greater accuracy in results. Some pupils
like arithmetic, others like algebra, still others like geometry ;
the change is helpful in keeping up interest. The study of
geometry forces analysis at every step and stage ; consequently
written problems and problems to be stated have no terrors
for those who are taught in this way.
For several years mathematical associations have urged that
all work should be based upon the equation. In accordance
with this view we have made the demonstrations in this book
largely algebraic, thus making the demonstration essentially
a study in simultaneous equations.
In this method of treatment, we have found it advantageous
not to hurry the work. Pupils gain power, not in solving
many problems, but in analyzing and t\io?o\x^t&3 xaAwafcaxs^.-
ing the principles of a few.
iii
iv PREFACE
Book I covers straight line geometry to proportion, and
algebra through fractional equations; it is intended for one
year's work.
Book II completes straight line geometry, takes up the study
of the circle, and on the algebraic side takes the student
through simultaneous quadratics.
We are indebted to many who have offered suggestions and
practical problems, and especially to Carlotta Greer of Cleve-
land Technical High School, and Professor Kenneth G. Smith
of the University of Wisconsin ; also to those who so kindly
read the proof sheets. We invite criticisms and suggestions
from teachers who are interested in progressive work along
these lines.
FOR THE TEACHER
Reviews in mathematics are always necessary. This is
especially true in this text, which combines different branches
of mathematics.
In teaching the text keep in mind that in geometry as well
as in algebra problems are solved by means of equations. The
equation is the principal tool used. To use the equation
method successfully, the Hilbert notation, a small letter for
angle values and for line values is essential.
In lettering a figure, begin at the lower left-hand corner and
read counter-clockwise. This gives pupils an idea of directed
lines, and makes possible the correct drawing of the figure
from description.
Emphasize the manipulation of quantities by means of fac-
tors and the use of methods of indication until there is no
longer hope that the factors may disappear through division
(see p. 162).
Do a large amount of the work orally, and do it so often
that the pupil knows what he is doing, and why he is doing
it. No pupil should use pencil and paper to find prime factors
of (24) 2 , (12) 6 , 9 • 27, or to find the product of 18 . 17.
Note that many demonstrations have been put into the form
of a set of simultaneous equations, the solution of which pro-
duces the desired equation.
Insist that pupils study all illustrative work, rules, and
instructions before beginning the examples of an exercise.
Teach pupils to use the Index, also the groups of theorems
and constructions found on pages 175-179.
Book I is considered one year's work. Book II is fspt^ka
second year.
v
TABLE OF CONTENTS
BOOK I
CHAPTER PAGE
I. The Number System 1
Addition . . 3
Prime Factors 5
Oral Control of Number 8
II. Equations . . . ... . . . .10
Problems 13
III. Positive and Negative Numbers .... 15
The Four Fundamental Operations . . . 16-27
Supplemental Applied Mathematics .... 27
IV. Inequalities . . . 33
Simultaneous Equations 35
Problems 43
Supplemental Applied Mathematics .... 44
V. Lines, Angles, Triangles . . . . .47
Supplemental Applied Mathematics .... 72
VI. Polynomials, Multiplication 76
Polynomials, Division 81
Review 86
Supplemental Applied Mathematics .... 88
VIL Graphs, the Algebra of Lines. Parallels and
•their Uses 92
Quadrilaterals 109
Polygons V\fe
Supplemental Applied Mathematics . ^^
vu
viii TABLE OF CONTENTS
CHAPTER PAOB
VIII. Products and Factors 122
(a + 6)(a-&) 122
(a + 6) 2 124
x 2 + kx + c 126
Common Factors 127
ax 3 + bx + c • 129
a 8 ±6 8 131
a: 4 + bx 2 + c 2 133
Solutions by Factors 137
Supplemental Applied Mathematics .... 139
IX. Fractions 143
Multiples . . . 147
Addition . . . 148
Multiplication 153
Division . 155
Equations • . . 157
Supplemental Applied Mathematics . . . .163
X. Proportion 170
List of Constructions 175
List of Theorems 176
Index 180
SECONDARY-SCHOOL
MATHEMATICS
BOOK I
CHAPTER I
The Number System
1. Our number system is a decimal one. Ten units of one
order make one of the next higher. One tenth of any digit
makes a digit of the next lower order. The digits are 1, 2,
3, 4, 5, 6, 7, 8, 9. All numbers are made up of these digits
and their position is often indicated by the introduction of one
other symbol, 0, known as nought, cipher, or zero; e.g., 16 is
equal to ten l's added to six Vs. That is, 16 = 10 + 6, the
position of the 1 being indicated by the 0.
The numbers expressed by these digits themselves are often
the niultiples of other digits. For example, 4 = two 2's, or
2 • 2, where the • indicates multiplication.
6 = 3-2, 9 = 3-3.
The numbers 1, 2, 3, 5, 7 are the prime digits. A prime num-
ber is a number whose only integral factors are itself and unity.
EXERCISE 1
Write the following numbers in such a way that their deci-
mal composition will appear :
1. 145. 8. 511.
145 = 100 + 40 + 5.
2. 223. 4. 987. 6. 227. 9. lftV.
3. 448. 5. 999. 7. ft6&. "\». A£fc\&.
2 MATHEMATICS
2. In the product of two or more numbers, any one of them
or the product of any number of them is a factor of the given
product. 2 • 3 • 5 = 30. Then, 2 is a factor of 30. 2 . 3 or 6
is also a factor of 30.
3. A term is a number whose parts are not separated by the
plus (+) or minus (— ) sign.
In the expression,
10 + 6
10 is a term. 6 is also a term.
10 + 6 is composed of two terms.
4. A binomial is an expression of two terms.
A trinomial is an expression of three terms.
A quadrinomial is an expression of four terms.
An expression of two or more terms is also called a poly-
nomial.
100 + 60 + 3 is a trinomial.
5. It is often necessary to represent a number by a letter or
a combination of letters. Such letters may represent either
unknown numbers or those supposed to be known numbers.
This kind of notation is used in general arithmetic or algebra.
E.g., n may represent any number, likewise any letter or
combination of letters and figures may be considered a number.
a + b + c is a trinomial number (§ 4), or the sum of three
numbers a, b, and c. In arithmetic it is possible to express
such a sum as a single number.
Thus, 2 + 5 + 8 = 15.
In algebra, this is not possible unless the terms of the expres-
sion are alike or similar.
6. Similar Terms are terms which differ in their coefficients
only, e.g., 5 • x, 6 • x, a • x, b • x.
THE NUMBER SYSTEM
7. Any factor (§ 2) of a number is the coefficient of th
maininer factors.
remaining factors.
Thus, in a • x, a is the coefficient of x.
in 2 • 3, 2 is the coefficient of 3.
in 2 • 3, 3 is the coefficient of 2.
in 2 • a ■• b, 2 • b is the coefficient of a.
in 2 • a • 6, a is the coefficient of 2 • 6.
in 2 • a • 6, 2 is the numerical coefficient of a • 6.
in ax, a is the literal coefficient of x.
When the product of a number of figures and letters is to t
written, the multiplication sign is usually omitted.
Thus, 2 • a • b is written 2 ab.
8. 2a + 3a + 7aisa trinomial consisting of similar tern
(§ 6). These terms may be united into one term by findin
the sum of the coefficients.
Hence, 2a + 3a + 7a= (2 + 3 + 7) a = 12a.
This is the same operation as that in arithmetic when one fine
the value of
2 f t. + 3 ft. + 7 ft. = 12 ft.,
and is brought still closer to arithmetic when one remembe:
that only like numbers can be added.
Similarly, 15 ab — 3 ab + 7 ab
means that 3 ab is to be subtracted from 15 ab, and 7 ab adcU
to this difference.
Hence, 15 ab — 3 ab + 7 ab = 19 ab.
Ex. 1. Find the sum of 20 xy + 4 xy — 7 d.
20xy + 4:xy-7 d= (20xy + ±xy) -7 d = 24 xy-1 d.
Ex. 2. Add 5 a 2 + 3 a& + 4 b 2 and - 4 a 2 -2 a&-4 & 2 .
ft /»2 j_ a aIi a. 4. w
4 MATHEMATICS
EXERCISE 2
Find the sum of the following .
1. 21x + 9x — 4:X + 3x>-8x.
21 a: + 9a? = 30 a
30x-4:X=:26x
26x + 3x = 29x
29s-8a? = 21a;.
This work is all to be done mentally, only results of each
addition being given.
2. 5m-4m + 6m-2m.
3. 8 xy + 3 xy — 2 d.
4. 8a + 46, 4a-26.
5. 16a 2 + 8a& + 56 2 , 5a 2 -3a& + 26 s .
6. 21x* + 22xy + lltf, -Sx* + 2xy-9y*,
and 7 a 2 -11 xy-ltf.
7. 24a 2 +48a& + 246 2 , - 23a 2 - 47a& + 23& f ,
and a 2 + 2 ab — b 2 .
8. 14 c 2 + 21 co* + 10 a* 2 , _ 9 c 2 - 12 ca* - a* 2 ,
and -5c 2 -9cd-9d 2 .
9. 3-19 + 2.19 + 5.19.
10. 3-27 + 2.27-4.27.
U. 14 • 18 + 25 • 18 - 16 . 18 - 12 . 18.
12. 41-63-27.63-12.63.
13. Express 27 as a binomial.
14. If x is the digit in tens' place and y in units' place,
express the number as a binomial.
15. Express 47 as a binomial.
is. Express 648 as a trinomial.
THE NUMBER SYSTEM 5
17. If hundreds' digit is x, tens' digit y, units' digit z,
express the number as a trinomial.
18. If the digits of example 17 are reversed, express the
number.
9. We have considered the decimal phase of our number
8y stem ; the prime factors are of equal importance.
The prime factors of 15 are 3 and 5.
45 = 3 • 3 • 5 or 3 2 • 5, where the 2 indicates the number of
times 3 occurs as a factor.
EXERCISE 3
1. Learn the following squares :
1 2 ,2 2 , 3 2 , 4 2 , 5 2 , 6 2 , 7 2 , ...30 2 .
2. Learn the following cubes :
I 8 , 2 s , 3 s , 4 s , 5 8 , 6 s , 7 8 , 8 8 , 9 s , 10 3 , 11 s , 12 3 .
10. Literal monomials may be separated into factors.
Thus, a 8 =(*•«• a.
a 2 6 = a • a • 6.
(a&)(a&) (a&) = (a&) 8 = a 8 ** 8 .
Similarly, (2 . 3) (2 . 3) (2 • 3) = (2 . 3) 8 = 2 8 . 3 8 = 6 8 .
Ex. Find the prime factors of 225.
225 = (15) 2 = (3.5) 2 ==3 2 -5 2 .
EXERCISE 4
Find the prime factors of the following :
1. 18,27,24. 6. (60) 2 .
2. (18) 2 ,(22) 2 . 7. 484.
3. 361,520. 8. (36/.
4. 9 s , (27) 2 , 729. 9. 625.
5. (12) 2 , (12/. 10. 225-12.
6 MATHEMATICS
11. An expression that is a factor of each of two or mo:
expressions is said to be a common factor of them.
Thus, 15a 2 6 = 3 • . 6 . a • a • b.
26a 2 6 = 6 2 -a.a- 6.
10 cfic = 2.6-aa«a-c-
5, a, and a are the factors common to each of the numbers 15 a' 2
25 a% 10 a*c.
EXERCISE 5
Find the factors common to the following :
1. 144,729. 6. 84 z 8 , 54 afe, 9.
2. 225, 5 4 . 7. 3125 a 5 , 625 a 4 , 125 a 8 .
3. (2 a) 8 , 24 a*c. 8. 243 a 5 6 10 , 162 a% 135 b\
4. 75 arty, 45ary 2 . 9. a A bPc?z, a 3 b A cz\ alPcx 7 .
5. 361 a 4 6*, 38 a 2 &, 114 a 3 & 3 . 10. 75 c 2 d, 125 cd 8 , 224 ad.
12. What does ofy mean ? How many factors are there i
the expression ? In arty, let x = 3, y = 5. What is the result
EXERCISE 6
In the following expressions substitute x = 1, y = 0, 2? =
a = 3, 6 = 4, and compute the value of the result :
1. 3a 2 6 8 = 3.3 2 .4 3 = 3. 9- 64 = 1728.
2. 2a# + 4y 2 + 7af% = 2.1. + 4. 2 + 7- 1 2 . 5
= +0 +35
= 35.
3. aar^z. Q y
4. 35x + 3y-6z. *
5. 10834 xyz. 10 y + Bx
z
6. (5a + 2b)x.
7. 2a(3& + 4*). u - (3f + Sx)(jf + Bx).
fi 7z(2x + by). 12. (fc-a)a;.
THE NUMBER SYSTEM 7
13. (b-a + y)x. 15. (3a + 2b + 2z)*.
14. An*. 1 6 . (i5 a _ 3 & _ 6 ^s #
13. The parenthesis as used in these examples denotes that
the quantities inclosed are each subject to the same operation.
Thus, (x + y)z means that the sum of x and y is to be multi-
plied by z, or that both x and y are to be multiplied by z and
the sum of the products taken.
(2 + 3) 2 indicates that the sum of 2 and 3 is to be " squared,"
i.e., used twice as a factor.
(a + b) + (c + d) is read : the sum of c and d is to be added
to the sum of a and b, or the sum of a and b plus the sum of
c and a\
Likewise, (a + b) — (c + d) indicates that the sum of c and d
is to be subtracted from the sum of a and b.
The forms of parenthesis are ( ), the brace \ }, the bracket [ ],
and the vinculum . The vinculum is seldom used.
EXERCISE 7
1. What does 12 x 2 mean ?
2. What does (12 x) 2 mean?
3. What does 12 (x) 2 mean ?
Perform the indicated operations :
4. (18a+12a)+(5a+2a). 9. 6(a + 3a) - 2(a + 2a).
5. (21a+2a)-(6a+3a). 10. 4[5(a+6a)].
6. (21a-2a) + (6a-3a). U. 6 [2(5 a- 3z)] -2 [4(2 a?-*)].
7. (21 a -2a)-(6a- 3 a). 12. (5<c + 18y) - (2y + 3y).
a 3(6s-2) +4(2x-l). 13. (8x + 12)| V^^-^V^tSY
8 MATHEMATICS
Oral Review
The area of a rectangle equals the product of the base and
altitude (length and breadth). Find the areas of the following
rectangles :
Length Breadth
1. *17" 6"
17-6 = (10 + 7)6 = 60 + 42
= 102.
2.
18'
6'
3.
18"
7"
4.
19'
V
5.
20'
9'
6.
22'
8'
7.
15"
12"
15-12 = 15(10 + 2) = 150 + 30
= 180.
8. 16' 12'
9. 18" 12"
The area of a triangle is equal to one half the product of the
base by the altitude. Find the areas of the following tri-
Length
Breadth
10.
19'
15'
u.
21'
24'
12.
17'
18'
13.
15'
13'
14.
12"
24"
15.
32"
21"
16.
19'
28'
17.
31'
22'
18.
106"
12"
19.
115"
12"
20.
106'
21'
21.
112'
15'
22.
1024'
4'
angles :
Base
Altitude
Base
Altitude
Base
Altitude
1. 16
10
6. 36
18
11. 37
19
2. 14
7
7. 44
22
12. 18
25
3. 24
8
8. 46
33
13. 25
16
4. 32
10
9. 48
24
14. 32
16
5. 24
12
10. 21
15
15. 17
21
* " indicates inches, ' indicates feet.
THE NUMBER SYSTEM 9
The area of a circle is equal to 3.1416 (Abbreviation is it,
pronounced pi) times the square of the radius.
Indicate the areas of the following circles :
(D represents diameter ; E, radius ; A, area.)
16. D = 12, A = tt'6 2 = 36it. 22. B = 27,A =
17. 22 = 18, ^ = 7r-18 2 = 3247r. 23. D = 28, A =
18. B = 20,A = 24. D = B6, A =
19. J9 = 38, A= 25. Z) = 58, A =
20. D = 46, A= 26. R = 26,A=:
21. 72 = 17,^1 = 27. 72 = 16, ^1 =
Find the side of a square whose area is :
28. 729 30. 900 32. 361
29. 841 31. 441 33. 529
i
Find the edge of a cube whose volume is :
34. 729 36. 64 38. 343
35. 512 37. 1331 39. 216
The area of a trapezoid is equal to the product of one half
the sum of the-lower base (B), and the upper base (6), by the
altitude (a).
Find the areas of the following trapezoids :
B
b
5
a
6
(10 + 5)6
: (10+ 5)3 =
45.
40. 10
2
B
b
a
B
b
a
41. 18
10
4
47. 18
8
8
42. 22
16
8
48. 16
14
9
43. 24
12
15
49. 15
13
14
44. 13
13
16
50. 24
15
6
45. 14
14
5
51. 22
11
8
46. 9
18
8
%
52. 15
17
17
Is there
anything
unusual about
the
trapexeAAs ycl ^sasrc^Rs*
44 and 45
•
CHAPTER II
Equations
14. An equation is a statement that two quantities are equal.
The sign of equality is =.
Thus, x + 3 = 5 is read x plus 3 is equal to 5.
15. To solve an equation is to find the value or values of
some letter involved in the equation which will satisfy the
given equation.
16. When a number substituted for some letter in an equa-
tion makes the sides of the 'equation identical, the equation is
said to be satisfied.
A number which satisfies an equation is called a root of the
equation. The number is also said to be a solution of the equation.
Ex. a> + 3 = 5. (1)
Substitute x = 2 in the equation.
Then, 2 + 3 = 5.
The two sides or members of the equation are the same or
identical.
The number on the left of the sign of equality is called the
first member or side. The number on the right is the second
member or side. Thus, in x + 3 = 5, x + 3 is the first member,
and 5 is the second member.
17. The kinds of equations that concern us at present are :
The equation of condition.
The identical equation or identity.
The geometric equation.
18. An equation of condition is an equation that is satisfied
only by a definite set of values.
10
EQUATIONS 11
E.g., in x + 3 = 5, x = 2 is the only value which can be found
for x, which is a root of the equation, x + 3 = 5 is therefore
an equation of condition, the condition being that a; must equal 2.
19. An identity is an equation which is always true for any
specified values of the letters involved in it.
E.g., 2 a = a + a is true for any finite value of a.
20. The geometric equation is an equation of two geometric
figures. In general, the algebraic equation (§§18, 19) is assumed
to be true, and if its roots satisfy it the statement of equality
is verified. The geometric equation must usually be proved
to be true (§ 70).
21. The operations used in equations are largely those of
addition, subtraction, multiplication, and division.
22. The laws governing the use of these operations are a set
of statements assumed to be true, and known as axioms.
23. The Axioms
1. If the same number, or equal numbers, be added to equal
numbers, the resulting numbers will be equal.
2. If the same number, or equal numbers, be subtracted from
equal numbers, the resultina numbers will be equal.
3. If equal numbers be multiplied by the same number, or
equal numbers, the resulting numbers will be equal.
4. If equal numbers be divided by the same number, or equal
numbers, the resulting numbers will be equal. It is not allowable
to divide by 0.
5. Any number equals itself.
6. Any number equals the sum of all its parts.
7. Any number is greater than any of its parts.
8. Two numbers which are equal to the same uuwJberc^ w \o
equal numbers, are equal.
12 MATHEMATICS
The Use of Axioms
24. Ex. 1. Solve 2 x + 3 = 9.
Subtract 3 from each member of the equation (Ax. 2).
2a: = 9-3.
or 2x = 6.
Divide each member by 2 (the coefficient of x) (Ax. 4).
Then, x = 3.
To verify this root, substitute 3 for x in the given equation.
2-3 + 3 = 9.
The equation is satisfied, hence 3 is a root.
Ex.2. Solve y+±± = 2l + !±.
6 3 2
Multiply both members of the equation by 6, the L. C. M. of the
denominators (Ax. 3).
K-
«l» + ^=V + i"
or 6y+ll = 4y + 15.
Subtract 4y from each side (Ax. 2).
Then, 2^ + 11 = 15.
Subtract 11 from each side (Ax. 2).
Then, 2 y = 4.
Divide both sides by 2 (Ax. 4).
And, y = 2.
Verify by substituting y = 2 in the grtoen equation.
a 32
Simplify each member.
12 + 11 _ 8 +15
_ 6~~ " 6 '
Hence, 2 is a root of the equation.
EXERCISE 8
Solve the following equations and verify each root:
2. 2i/+7y + 3 = 12. 2. 5»4-7=3a + 17.
EQUATIONS 13
5x lx Q 5 5. 43-2=6 (Ax.).
6 4 12
6. 7<c — 4 = a + 14.
4. 6* + i=»2* + |. 7 ' 8* + 2-3*-4 = 4*.
a 6u-4 = 3w + 8.
y+ 5 + 4 3 + 20*
10. The sum of two numbers is 9, and one is twice the
other. Find the numbers.
Let x = the smaller number,
and 2 x = the greater number.
2z + x = the sum of the numbers.
9 = the sum of the numbers.
Then, 2 x + x = 9 (Ax. 8),
or 3 x = 9.
x = 3, the smaller number.
2 a = 6, the greater number.
i
11. The difference between two numbers is 24, and the
greater is four times the lesser. Find the numbers.
12. The sum of two numbers is 48, and the greater is
four more than the lesser. Find the numbers.
13. A number is composed of two digits. The tens' digit is
four times as great as units' digit, and the sum of the digits
is 10. Find the number. (Exercise 2, Ex. 14.)
14. A number is composed of two digits. The tens' digit is
three times the units' digit, and the number is 54 more than
the sum of the digits. Find the number.
15. The distance around a rectangle is 120 r . The length of
the rectangle is 10 r more than the breadth. Find the dimen-
sions.
14 MATHEMATICS
16. Two rectangles each have an altitude of 8', the sum of
their areas is 256, and the difference of their lengths is 12'.
Find length of each.
17. In a triangle ABC the area is 24, the
altitude (a) is 8. Find the base (6).
ia In a trapezoid ACDE, the area is
120; the lower base B is four times the
upper base b\ the altitude is 8. Find the
b bases.
19. The sum of the angles of a triangle is equal C
to 180°. The angle at A is 20° more than the yA
angle at B, and 50° more than the angle at C. ££- — AB
Find the angles.
20. In the triangle ABC, Z A is twice Z B, and Z C is equal
to the sum of Z A and Z B. Find the angles.
21. The sum of two numbers is 48 and one number is four
times the other. Find the numbers. (Solve mentally.)
In the following examples determine whether they are iden-*
tities or conditional equations, by substituting values for x.
If you find more than two values that satisfy the equation, it
is an identity.
22. x 2 -6x + 9=(x-3)(x-3).
23. ar^-f 12 x = x(x + 12).
24. 4^-9^4-7 = 3^-^-8.
25. 4s 2 -12a; + 9=:(2a;-3)(2a;-3).
26. 6x 2 -2x + 4: = 2x 2 + 10x-5.
27. 5x* + 6x + 4: = (5x-2)(x + 4:).
28. a 2 -12a; + 36 = (a;-6) 2 .
29. x 2 + 36x-j-Sl=:(x + 9)(x + 9).
CHAPTER III
Positive and Negative Numbers. The Four Fundamental
Operations
25. Positive and Negative Numbers. In addition to the num-
bers thus far used in computation, it is necessary in algebra to
extend the idea of number somewhat farther. In many prob-
lems the numbers involved seem to have an opposite sense.
For example: If a man has $500 and owes $300, the $300
opposes the $500. If a man is walking east, the opposite
direction to that in which he is going is west. North and south
are opposites. Temperatures above and below zero are oppo-
sites. To express this sense of opposition, positive (+) and
negative (— ) signs are used in mathematics. E.g., if toward
the right is +, then toward the left is — . If north is con-
sidered positive, south is negative. If assets are +, liabilities
are — .
EXERCISE 9
AC D B 1. If D is four units to
the right of 0, and C two
units to the left of 0, how far is it from C to D ? Does that
mean to you the difference between the positions of point C
and point D? If A is —3 units from 0, and B is +5 units
from 0, what is the distance from A to B ?
• 2. A man has $600 and owes $300. How much is he
worth?
3. A man has $ 500 and owes $ 700. How much is he worth ?
4. A man has $500 and owes $500. How muck v& ^&
worth ?
15
16 MATHEMATICS
5. Where on the Cleveland- Wooster railway line is a place
— 10 miles north of Cleveland ?
6. A man goes 5 miles north of Cleveland, then 9 miles
south. How many miles north of Cleveland is he? How
many miles has he traveled ? Draw a diagram showing his
route and his last position.
7. If to the right is positive, measure + 6" from a point A.
Call this point B. Measure — 9" from B. Call this point Z>.
Where is D with respect to A ?
8. Translate into English (a 2 — ft 2 ) 8 .
Write in symbols : The square of the result of the quo-
tient of the sum of (a) and (b) by 2.
9. The temperature at 6.00 a.m. is 4- 14° and during the
morning it grows colder at the rate of 4° an hour. Required
the temperatures at 9 a.m., at 10 a.m., and at noon.
10. Find the numerical value of the following, when a = 3,
6 = 5, c = 2, d = 4: a*b_atf
2c 2 4d 2 '
U. Translate into English f*±IN.
Write in symbols : the cube of the result of subtracting
the square of b from the square of a.
12. Find the numerical value of the following, when x = ^,
y=*,z = \: 4z 2 +(3y-2*) 2 .
Addition
26. In § 8, we found the sums of similar terms, but in each
instance the sum was positive. A negative sum may arise from
the addition of a positive and a negative number. For instance,
in example 6, exercise 9, we are adding — 9 miles to + 5 miles.
The result is — 4 miles. That is, the addition of a negative
numder to a positive number tends to lessen the numerical value
FOUR FUNDAMENTAL OPERATIONS 17
of the sum, annul it, or change it from positive to negative. If a
man has $400 and owes $400, the sum of his assets and liabil-
ities is 0. If he has $400 and owes $700, the sum of his
assets and liabilities is — $300. That is, he owes $300 more
than he has assets. The sum of two negative numbers is nega-
*
tive. It is seen in these results that when the sum of a positive
and a negative number is found, the result takes the sign of
the greater absolute value.
27. The absolute value of a number is its value regardless of
sign. For example, the absolute value of — 6 is 6.
28. If no sign is placed before a number it is regarded as
positive. A negative sign must never be omitted.
EXERCISE 10
Find the sums of the following:
1. 4-5 2. -f5 3.-5
4.-5
4-3 -3 4-3
-3
5. 3 a 2 and — 4 a 2 .
6. 7 ab, — 5 ab.
7. — 8 x, — 5 x.
8. 9a 2 , -6 a?, - 3 a 2 , - 12 a 2 , - 5 a 2 .
9. 3 a, — 5 a, 4- 6 a, — 4 a.
*10. 7a 2 , 4<c, 4- 3a 2 , +2x, + x.
11. 12x*y, 6xy 2 , 4afy, - 2afy, 4- llmf.
12. 15^, - IBx 2 , + 15x, 4-5Z 2 , -5a?, -9ar*, -9ar>, -9aj.
13. 4a?4- 3y — z, — 2x + y 4- 4z, — x — 3y — 2z.
14. 5 (a 4- b), - 2 (a 4- b), 6 (a 4- &), - 9 (a 4- b).
* Write similar terms (§6) in the same column. Make as many columns as
you have different kinds of terms, forming the whole Into owe ^t^V>\&\xv N \^
using 4- and — signs. (See example 2, \ %.)
S
18 MATHEMATICS
15. 4 (x + y) - 3 (x - 2 y), - 3 (a + y) + 2 (x - 2 y),
2(s + y) + (x~2y).
16. 2 (a + &) 2 - 6 (a + b) + 1, - 5 (a + fc) 2 + 3 (a + 6) - 6,
( a + &) 2 - ( a + b) + 2.
18. fa-fft + fc, ^a-tft-fc, -fa + f&- A*
Find the value of the following sums when x = J, y = + £,
« = i,a = -2,c = |,6 = 2:
19. £a + i&-c, + a -J6-|c, 5a — £& + 2c.
20. 2(a + b) 2 - 3(6 - c) 2 + (a - 6), - 500(a + 6) 2 + 6(ft - c) 2
+ (a-6).
21. 5 xy — 5 afy — 5 scy 2 , ^ ay + f afy
22. 7(s + 2y)-7(a;-2y), -31(x+2y) + $(x-2y).
23. 12 yz — Sxy + Ja + ffcc.
24. $a-ffc + fc, + fa-£fc + 25c.
In each of the following examples, add corresponding mem-
bers of the two equations to find x :
25. x -f y = 8,
a? — y = 4 . When a? is found, can you find y?
26. 2a; + 3y = 8,
a? — 3y = — 5 . Find y also.
27. 4a?-2y= 2,
3a? + 2y = 12 . Find y also.
2a Verify your results in examples 25-27 by substituting
the values found for x and y in the given equations.
Subtraction
29. (a) What number added to 9 gives 7 ?
(d) What number added to 9 gives 11 ?
FOUR FUNDAMENTAL OPERATIONS 19
(c) What number added to 9 gives ?
(d) What number added to 9 gives — 12 ?
In each of the above examples we have the sum of two num-
bers and one of the numbers given to find the other number.
The Minuend is the sum of two numbers.
The Subtrahend is a given number.
The Difference is a required number when the minuend and
subtrahend are known.
Subtraction is the process of finding what number added to
the subtrahend produces the minuend.
In example (a), 7 is the minuend, 9 is the subtrahend, — 2
the difference. In subtraction, tne sum of the subtrahend and
the difference must equal the minuend. This fact enables us
to check our result.
EXERCISE 11
1. Subtract — 3 from 5.
Here 5 is the sum of the numbers. — 3 is one of the numbers. Our
problem is : What number added to — 3 gives 5. This number is evi-
dently 8. That is, the difference is 8.
Check : 8 + (— 3) = sum of the numbers.
5 = sum of the numbers.
The written work stands :
5
-3
2. From — 8 a take 5a:
8
- 8a
4- 5a
-13 a
Perform the following subtractions :
3. Itfy 4. 25a 5. 13a; 6. —25a 7. +25a;
-3afy 13a 25x -\2>x -V^**
20 MATHEMATICS
8. From — 13 ab take 24 ab.
9. From a -f- b -f c take a — b — 2 c.
(Note that there are three subtraction examples in this example, one
for each column.)
10. Subtract 8 x — 2 y -f- z from 10 x — y — 3 2.
11. Subtract 2a 2 — 3a + l from 5a 2 - 3a- 1.
12. Subtract 10 a 2 + 5 ab - 9 6 2 from 2 a 2 - 10 a& + 8 b 2 .
13. From or 3 + 3 or 2 -f 3 a; + 1 take #* + 2o 2 + 2a; + 1.
14. From ar< + 3:c 2 + 3a:+l take X s - 3 a 2 + 3 x - 1.
15. From a 2 + 2 ab + 6 2 take a 2 - 2 a& + 6 2 .
16. From 5 x 2 take 5 y 2 .
17. From 6o 2 -5a + 4 take 12a; ? -9.
18. FromSa^take - 4o 3 -f-3a 2 - 2<c + l.
19. Subtract a from 0.
20. Subtract 9 a? 2 + 9 y + 9 from 0.
Note that in each of the above examples the difference is
the same as if we had changed the sign of the subtrahend and
added the result to the minuend.
We may then use the following rule for subtraction. Change
the sign of each term of the subtrahend and proceed as in addi-
tion, (The change of sign must be made mentally.)
EXERCISE 12
1. What must be added to9a 2 + 9i/ + 9to give ?
2. From x 2 + 4 x + 4 take x* — 4 x 4- 4.
3. From the sum of a 2 -f- 2ab-\-b 2 and a 2 — 2 ab + b 2 take
the difference between a 2 + 2 ab + b 2 and a 2 — 2 ab -f- b 2 .*
* In this text the difference between a and b means the remainder obtained
by subtracting b from a.
POUR FUNDAMENTAL OPERATIONS 21
4. Subtract 2 a* + a 2 6 + a& 2 + 2 6 8 from the sum of a 8 + 8 a*b
+ 3 a& 2 + ft 3 and a 3 - 3a 2 6 + 3a& 2 - 6 s .
5. Subtract x* -f- 3 a; — 4 from a 3 .
6. Subtract 6 + c from a.
7. Subtract 4 a 2 — 7 a — 9 from 0.
8. What shall we add to 7 a 2 — 12 a; + 5 to produce ?
9. [5aj 3 -(3aj-f-2)]-[6a 8 4-(4jc-ll)].
In examples of this type perform the indicated operations
one at a time, beginning with the inner parenthesis. First sub-
tract 3 x -f- 2 from 5 x 3 , then add 4 a: — 11 to 6 x\ This gives
[5x* - 3« - 2] - [6a 3 + 4 a- 11].
Subtracting the second trinomial from the first, we have
_ar8_7a;-f 9.
Translate each example into English before solving :
10. {5a-[2a+(4a + l)]j.
11. [12« 2 -{-7^-(5aj 2 + 6a;-3)}].
12. [6(5a + 3) + 5(2a + 7)]-(6a + 53).
13. [25m-(2wt + 3)] — [-10 m— (6 m -7;].
14. [a?-(Sa^f-S*/-^)]-[^ + (-3^y+3^-/;].
15. (12a + 36 + 2c)-(5a-26 + 6c) -(10a -&_0c).
16. 4(* + y)-[6(*-y;] + (*-2y).
17. There are two numbers, z and y 9 whose sum i* 17 and
whose difference is l,x being the greater number. Find the
numbers.
By the condition*,
* + F=17
0)
r. — y= 1
f2;
Adding (1) and f2),
% = <*.
Sabtzaetmg r?> frcnw 1>,
22 MATHEMATICS
18. Given two numbers, x and y, such that the second num-
ber added to twice the first is equal to 12, and the difference
between twice the first number and the second number is 8.
Find the numbers.
19. 5x + 2y = 19, (1)
5x- 2y = ll . (2) Find x and y.
20. x +3y = 16, (1)
x-3y=-U . (2)
21. 2x + 3y = 17, (1)
x — 3y= -—14 . (2) After x is found, obtain the
value for y by substituting the value of x in equation (1).
22. Find the value of the difference between 5 x*+ 3 x*y— 5 a 8
and 2 x* — 3 afy — 2 a 2 , when » = 5, # = £, a = 3.
Multiplication
30. In addition we learned that
5 + 5 + 5 + 5 = 54 = 20
Also that
(-5) + (-5) + (-5) + (-5) = (-5)(4) = -20.
This last shows that the product of a negative number by
a positive number is negative.
In arithmetic, it does not matter in what order the factors
are used, e.g., 5*3 = 3*5.
We shall assume that this law holds true in algebra.
Then, (- 5) (4) = (4) (- 5) = - 20.
That is, the product of a positive number by a negative number
is negative. It seems then that multiplication by a negative
number gives to the product a sign opposite to that of the
multiplicand.
JEx. (-8)- (-5) =+40.
FOUR FUNDAMENTAL OPERATIONS 23
31. These rules for signs follow:
The product of two numbers of like sign is positive.
The product of two numbers of unlike sign is negative.
32. Since division is the inverse of multiplication, the same
sign rules hold ; namely,
Like signs give phis and unlike signs give minus.
Namely, when the terms have like signs, the quotient is posi-
tive ; when unlike, the quotient is negative.
Division may be regarded as an application of the method
of subtraction.
Ex. Divide 28 by 7.
This is equivalent to successively subtracting 7 until no remainder or a
remainder less than 7 remains.
28-7 = 21, 21-7=14, 14-7 = 7, 7-7 = 0.
Which shows that 7 is contained in 28 four times without a remainder,
or that 7 is an integral factor of 28.
Regarding division as the inverse of multiplication is the
more general method. With this understanding we have the
following definitions.
33. Division is the process of finding one of two factors
when their product and one of the factors are given.
The Dividend is the product of the factors.
The Divisor is the given factor.
The Quotient is the required factor.
It is evident from these definitions that the product of the
divisor by the quotient is equal to the dividend. By the use
of this principle the accuracy of the result of the division may
be verified.
Give sign rules for addition, subtraction, multiplication, and
division.
24 MATHEMATICS
EXERCISE 13
Find the product of the following (first determine the sign
of the product) :
1. (4)(-3). 4. (+10)(+10).
2. (-9)(4). 5. (-10)(-10).
3. (-9)(-8). 6. (-7)(+9).
Find the following quotients :
7. (+12) -J- (+4). 10. (-20) -K-4).
8. (-12) -s- (-3). 11. (-112) -(16).
9. (_i5)_5_(+5). i2. (144) + (-18).
13. The area of a rectangle is 212, the base is 17. . Find the
altitude.
14. The area of a triangle is 180, the base is 20. Find the
altitude. Can you construct the triangle ? Why ?
15. The area of a rectangle is 24, the altitude is — 4. What
is the base ? How do you account for these negative results ?
Draw this rectangle. (§ 25.)
34. In multiplication it is convenient to show the number of
times a quantity is used as a factor by means of a symbol
placed at the right and above a quantity.
This symbol showing to what power the number is to be
raised is the exponent of the number.
Ex. 1. x-x-x = x 3 . (2 x)(2 x)(2 x)(2 x) = (2 x) 4 = 16 x\
Similarly, x • x • x ••• n factors = x*.
Ex. 2. x 2 'X 4 = x-x-x»X'X*x = x*.
And x w • x m = (X'X-x ••• n factors) {x • x • x ••• m factors)
= X'X'X ••• m + n factors = x m + n ,*
Then in multiplying two like letters give to their product an
exponent equal to the sum of their exponents in multiplicand
and multiplier.
* ... is read, " and so on to."
FOUR FUNDAMENTAL OPERATIONS 25
We see that multiplication is simply combining the factors
of the multiplicand and multiplier into one product ; e.g., 12 • 18
means (2 2 • 3) (3 2 • 2) or 2 8 • 3 8 , the product containing all the factors
of both multiplicand and multiplier. Similarly, 24 a 2 b • 15 a 8
= 2 8 • 3 • a* • & • 3 • 5 . a 8 = 2 8 • 3 2 • 5 • a 5 • b = 360 a 5 &.
35. Since division is the inverse of multiplication, the ex-
ponent rule for division is the reverse of that in multiplication.
In dividing a letter by the same letter having the same or a
different exponent, give to the quotient an exponent equal to
the exponent of the dividend minus the exponent of the divisor.
Ex. 1. Divide X s by x 2 .
a; 8 = x • x • x,
x 2 = x • x.
Then, (x • x • x) -*- (x • x) = x,
or X s -*- x 2 = x 3-2 = x 1 .
When the exponent is 1, it is not expressed.
Thus, x 8 -T- x 2 = x.
Ex. 2. Divide a u by a 12 .
a 15 -4- a 12 = a 16 " 12 = a 8 .
EXERCISE 14
Eind the following indicated products :
(Translate each example into English before solving.)
1. X s 'X 3 . 8. (12 <*) • (- 4 c 5 ).
2. 15 a 2 . 5 X s . 9. a*b-a 4 b 2 .
3. 8 m- 18 m 4 . 10. -15a 2 & 2 • 21 a 3 b 5 .
4. (15a) 3 (2a). 11. (- 13 aW)(- 13).
5 - + 2/) 2 + y) 3 - 12 - (IS <M)(- 19 Jtf).
6. 12(a + 6) • 3(a + &). 13. (27 ab 2 xy)(- 14 a 5 &c).
7. -19(a-6) 4 .15(a~6) 3 . 14. (^-2,1^%)^-^^^
26 MATHEMATICS
15. 12(a + b) by 12(a + b) % by 12(a + ft) 8 . What power of
2 is in your product? What power of 3? What power of
(a -f- b) ? What then are the prime factors of your product ?
Indicate the prime factors of these products :
16. (18a 3 b*)(18ab*).
17. (- 24p 2 g 4 r)(- 24^Yr).
18. 32(a + x) 5 by 64(a + a) 6 .
19. 27(c + d) s by 9(c + d)\
20. 125(a; - yf by 25(a - y) 2 .
Find these quotients and verify your results :
21. a^-r-ar 5 . 27. 729 c 4 x 7 -*- 9 ex 3 .
22. a 4 -*- a?. 28. (c— d) 10 -h (c — d) 4 -
23. 72z fi -j-9» 2 . 29. 27(a + b) 3 -5- 9(a + b) 2 .
24. 441 a 6 -*- 21 a 8 . 30- a s b 3 <? + a 2 be.
25. (-24 a?) +(-3 a 1 ). 31. 2 5 . 5 8 . 3 2 -s- 2 2 . 5 . 3.
26. (576a 7 6 4 )-s-(-24a 5 6 2 ).
Are examples 30 and 31 similar ?
32. Divide 5 4 . 3 5 . 2 4 a 7 by by 5 3 . 2 4 a 7 b«y 2 .
Factor the dividend and divisor. Then divide as in ex-
ample 32 :
33. 162 afy -s- 54 xy- 34. — 135 cV h- 15 ex 5 .
35. - 210 «*yV -h - 14 aty 8 .
36. The area of a rectangle is 12S(x + y)(c + d), its altitude
is 8(c -f- d). What is its base ? What are its dimensions if
a? = l, y = 2, d=4, c= -3?
37. The area of a rectangle is 384 rfyc 2 , its altitude is 16 a^c.
Find its base. What are its dimensions if oj = — 2, y= — 1,
c = — 2 ? Also if x = 2, y = 1, c = 2 ? Draw these rec-
tangles, measuring from the same starting point for each.
FOUR FUNDAMENTAL OPERATIONS 27
3a The base of a triangle is 27 aV, the area is 243 a A <?.
Find the altitude. What are the dimensions when a =* — 3,
36. Give the sign rules for addition, subtraction, multipli-
cation, division.
Give the exponent rules for multiplication and division.
Supplemental Applied Mathematics
A decimal fraction is a fraction whose denominator is a power
of 10. This denominator may be written, or simply expressed
by the relative position of the decimal point. The factors of
10 are 2 and 5. The factors of 100 are two 2'a and two 5%
that is, 100 = 2 2 -&. Likewise 1000 = 2*-5*. Every power of
10 is made up of an equal number of 2'b and 5's used an fac-
tors. Therefore, to reduce a common fraction to a pure dswimsjtf
one must multiply both numerator and denominator by such
number as will produce an equal number of 2 and 5 factor*
only in the denominator.
Ex. 1. Reduce f to a decimal*
« _ Two h'% are lacking. Multiplying both uuiaerzt/*r und
2 = am den«ninator by 5* or 25, we hare /^ c ; or, ezpreftttd d**;i jaiaJJy,
.75.
Ex. 2. Reduce to .1875 a common f ration :
1*75 _ «•# _ 3
2* •>* £*.,y K
3. Beduce to dwinals: |. f. / c . J J, J, £$-
4. A macliiiiifrt has a K^t of dri^s vy<xr)cA .1250; ,S#75;
.0625; .03125: -2125; Jf7^ If* does not rw/piy/M Xh*w **
readily as if they *r?re in fctLfc, Vj\Lt, Z&ii, *'A\h>„ \'*v*+*m
them to saeh itota.tico-
5. Reduce tl*e fVI>r»;:_/ t^« of 'JrJUt to t;,«r;r cU*%;:-feJ
equivalent*; 1-8, 3-31', ;*-K. 5-10, 7-lo\ SM0 ; 11-10, l^-t<; n
19-16.
28 MATHEMATICS
In all computation in science and shop practice the decimal
plays an important part.
Whether multiplication and division is carried on directly or
by means of tables, the computer must know&t a glance where
the decimal point is to be placed.
Multiplication of Decimals
Decimals are multiplied as simple numbers if one remembers
the decimal composition of our system and the part the posi-
tion of the digit plays in the formation of a number. In 145,
§ 1 and exercise 1, the 1 has 100 times the value it would have
if written in units' column where the 5 now is. The 4 has ten
times the value it would have if in units' column. Then, mul-
tiplication by a digit in hundreds' column has 100 times the
effect of multiplication by the same number in units' column.
A similar statement holds for digits in tenths' and hundredths'
columns, each move to the right decreasing the value of a digit
ten times.
In multiplication we begin at the left.
Ex. 6. Multiply 24 by 36.
24
36
720 '
144
864
We multiply first by 3. Since this figure is in tens' column, it has ten
times the value of a figure in units' column, and our product moves one
place to the left. That is, we are really multiplying by 30 units. The
units' column, not the decimal point, is the dividing line.
Ex. 7. Multiply 23.2 by 2.4.
23.2
2.4
46.4
9.28
65.68
FOUR FUNDAMENTAL OPERATIONS 29
Always keep the decimal points in the same vertical column.
When multiplying by 4, the multiplier was one place to the
right of units, and the product ^ as large as the product by the
same figure if in units' place. This shifted our product, 9.28,
one place to the right.
Always begin multiplication at the left.
Find the following products :
a 17-26. 17. (2.7)*. 26. .6425 (.0125).
9. 324 • 14. la (.27)*. How much multiplying
10. 324 . 324. 19. (.09) 8 . is necessary in ex-
U. 216-36. 20. 24.21 (.32). ample 27, if the result
12. 9 s - (3.1416). 21. 1.875 - 16.2. is correct to three
13. 16 s - (3.1416). 22. 1.875-1.62. decimal places ?
14. 24 . 62.5- (.92). 2a 1.875 (.162). 7n ' 4 -261 (.7854).
15. (.8) 8 . 24. 18.75(1.62). * 32.15 (.625).
la (1.2) 8 . 25. 1.112 (.99).
of Decimals
Here we must reverse our work of multiplication. The first
figure of the divisor and its distance from units' column deter-
mine the position of the decimal point with respect to the first
figure of the quotient. If the first figure of the divisor is in
tens' column, the quotient will be ten times as small (or one
tenth as much) as if the divisor were units. The first figure of
the quotient therefore moves one place to the right of the first
figure of the dividend.
Ex. 29. Divide 144. by 72.
02.
72. ! 144.
J44.
Note that 7 is not contained in 1. the first figure rrt V\\fc qp\cA\«-xv\.\ *W*<o
fore a zero is written tot the first figure of the qwAvent. Vauc*. ^* , ** slt *
30
MATHEMATICS
is to the left of an integer, it has no value and may be disregarded. Until
the beginner has the placing of the first figure of the quotient well in mind,
this zero should be employed.
Ex. 30. Divide 324. by 18.
18.
18.
324.
18 .
144.
144.
Ex. 31. Divide 2.446 by 3.2.
0.76 +
3.2
2.446
2.24
.206
.192
.014
Ex. 32. Divide 2.446 by .32.
07.6+
.32 | 2.446
2.24
.206
.192
Ex. 33. Divide 2.446 by 32.
.014
.076 +
32.
2.446
2.24
.206
.192
.014
Note that the position of the first figure of the divisor alone controls
the position of the first figure of the quotient.
Find the following quotients. (Use three decimal places if
the quotient is not exact.)
FOUR FUNDAMENTAL OPERATIONS 31
?4. 31.5 -=-24.6. 44. 251.328-s-3.1416.
35. 6.4-s- 1.6*. 45. 442 -5- .09.
36. 6.4 -5- .16. 46. 2.24-*- 22.4.
37. 6.4 -f- .016. 47. 361. -S-.19.
3a 6.4-^-16. 48. 5.29 -«- .23.
39. .225-5-15. 49. 7.29 -f- 2.7.
40. .225 -7- 1.5. 50. 7.29 -j- .27.
41. .225 -s- .15. 51. .0289-5-1.7.
42. 109.624 -s- 3.86. 52. .0289 -s- .17.
43. 125.664 -s- 3.1416. 53. .0289 + 17.
Locate by inspection the first figure in each of the following
quotients. (Remember that the number of places in the quo-
tient does not in any way depend upon the number of figures in
the divisor.) The position of the first figure in the divisor is
all that one needs consider.
For example, in dividing 8.432694 by .3419768321, the first
figure 2 takes the same position as if we were dividing by .3,
namely, in tens' column.
2 .
.3419768321 | 8.432694
54. 48.36579-5-4.6293251. 60. 34.76-5-38.
55. 4.836579-5-4.6293251764. 61. .026947321 + .41976384.
56. 4.836 -j- 4.6293251764847. 62. .178643791-5-2.9.
57. .2793 -*- .217398. 63. .243 -*- .0986432791.
58. .2793 -r- .0217398. 64. 2.43 -s- .986432791.
59. 31.84-5-309.7. 65. .0243-4-9.864327915.
66. Multiply 16346.2' by .00019, and show that the product
is in miles.
67. Divide 3 cubic feet by .00058, and show that the quotient
is cubic inches.
32 MATHEMATICS
68. Rolled oats requires If hours cooking on a range. If a
fireless cooker is used, 15 minutes cooking on* the range is suffi-
cient. How much fuel is saved by using the fireless cooker, if
8.6 cu. ft. of gas per hour are consumed by a gas burner, gas
costing 76^ per thousand cubic feet ?
69. Wheatena, or cream of wheat, requires £ hours cooking
on a range, or 15 minutes cooking on a range for a fireless
cooker. How much fuel is saved by using the fireless cooker,
gas burner and price same as in Ex. 68 ?
70. Corn meal mush should be cooked for three hours on a
range, or for 15 minutes if a fireless cooker is used. How
much fuel is saved by using a fireless cooker ?
71. Uncooked rice contains 79 % carbohydrates, while boiled
rice contains 24.4 %. How much carbohydrate is lost from one
pound of rice by boiling ?
72. Rice boils in 20 minutes, using a gas burner consuming
8.6 cubic feet per hour. Steamed rice is cooked on the same
burner for 5 minutes, and on the simmering burner for 45
minutes, the latter consuming 3.6 cubic feet per hour. What
is the difference in the cost of cooking ?
73. Rice swells 3 J times by boiling. If a recipe for pudding
calls for one quart boiled rice, how much uncooked rice should
be used ?
CHAPTER IV
Inequalities. Simultaneous Equations
37. The sign of inequality is > or < , the opening being
toward the greater number.
Thus 9 > 5 is read, " 9 is greater than 5."
4 < 7 is read, " 4 is less than 7."
— 5 > — 9 is read, " — 5 is greater than — 9."
38. Two inequalities are in the same sense when the first
member of each is greater or less, respectively, than the second
member.
39. If equal numbers be added to or subtracted from both
sides of an inequality, the resulting inequality is said to con-
tinue in the same sense. That is, the inequality sign is not
reversed.
Ex. 9 > 5. Subtract 3 from each side. Then
9_3>5_3
or 6 > 2.
40. If all of the signs of an inequality are changed, the in-
equality does not continue in the same sense.
That is, if all the signs of inequality are changed, the sign of
inequality must be reversed.
41. An inequality continues in the same sense after being
multiplied or divided by a positive number.
Ex. 1. 2a>10.
Dividing both sides by 2, we have
x > 6.
33
34 MATHEMATICS
Ex.2. 2x + y>8, (1)
.V = 2. (2)
Subtract (2) from (1) , 2 x > 6 (3) (§39)
Divide (3) by 2, x > 3 (4) (§ 41)
That is, to satisfy the above conditions, x must be greater than 3.
Transposition
42. In § 24 we solved an equation,
2<c + 3 = 9
by subtracting 3 from each side, by Axiom 2, then dividing
both sides of the resulting equation by the coefficient of the
unknown number.
By a method called transposition, it is possible to simplify
the first operation.
Ex. 1. Solve for x 7 mx + c = b. (1)
Subtracting c from each side, mx = 6 — c. (2)
Note that the equation (2) is the same as if we had removed -f c from
the first member of the equation (1) and placed it in the second member
with ite sign changed.
Such change is called transposition.
This rule follows : A term may be transposed from one mem-
ber of an equation to the other if its sign is changed.
Then, in solving, mx + c = 6.
Transposing c, mx = b — c,
and dividing both members of the equation by m, the coefficient of g,
m
Ex.2. oa?-f 7=13-2ax. (1)
Transposing the unknown term — 2 ax to the first side, and -f 7 to the
second side, we have ax + 2 ax = 13 — 7, (2)
3 ax = 6, (3)
oa a
INEQUALITIES 35
To verify this value of x, substitute the value of a; in (1).
or 2 + 7 = 13 - 4.
9 = 9. .\ is the abbreviation for hence.
Simultaneous Equations
43. In examples of exercise 8 and the last ones of exercise 10,
we note that in the former exercise one unknown number is
involved in the equation, while in the latter the values of two
unknowns must be found.
The method for the solution of equations in two unknowns
is to find some way of combining the two equations into one
equation containing one unknown, and solving this resulting
equation.
Such process is called elimination.
Two equations in two unknowns are necessary because one
equation in two unknowns has no definite solution, but has an
indefinite number of solutions.
E.g., in
x + y = 6, if x = 0, y = 6; as = 1, y = 6 ; x = 2, y = 4 ; x = S, y = 3 ; etc.,
any number of pairs of values satisfying the given equation.
44. Note that in this equation y changes value when x
changes. This is because x and y are so related that their sum
must always be 6. Such values must be given x and y that
the equation must be kept in balance ; that is, must be kept an
equation.
The unknown numbers which are subject to change of value
in an equation are often called variables.
45. The usual methods of elimination are addition or sub-
traction, substitution, and comparison.
46. First Method. Addition or Subtraction. In this \x\atWL
one or both equations are multiplied m sueta a.Tx^ctftfst ^foa^^^
36 MATHEMATICS
coefficients of the same unknown in both equations become
identical. One unknown number may then be eliminated by
addition or subtraction of the corresponding sides of the equa-
tion.
Ex. 1.
2s+ y = 9,
(1)
3x-2y = 10.
(2)
Multiply (1) by 2.
4se+2y = 18
(3)
Add (2) and (3),
3 x - 2 y = 10
(2)
Whence,
1x =28
(4)
and
x =4.
Substitute x = 4 in
0),
8 + y = 9.
(5)
Transposing (§ 42)
+ 8 in (5), we have,
y = 9 - 8,
or,
y = l.
Check these values
in (2),
3 • 4 - 2 • 1 = 10,
12 - 2 = 10.
Ex.2. Sx- 5y = 31, (1)
12x+13y=-15. (2)
Multiply (1) by 3, and (2) by 2,
24x-15y = 93 (3)
24 x + 26 y = - 30 (4)
Subtract (4) from (3), - 41 y = 123. (6)
Divide (5) by —41, the coefficient of y.
Whence, y - - 3. (6)
Substitute value of y in (1),
8x- 5 (-3) =31,
8 a: + 15 =31. * (7)
Transposing (§ 42) 8 x = 31 — 16.
8 x = 16. (8)
Whence, x — 2.
Check in (2), 12 . 2 + 13 (- 3) = - 15.
24 - 39 = - 15.
EXERCISE 15
Solve the following equations and verify results :
1. 3a-42/ = l, 2. 5* + 7% = 29,
4x + 3y = l$. ■ 5t-7u = t.
INEQUALITIES 37
3. 3i> — 521=12, 7. \x + %y=:i£,
8??+10g=-38. $a?-|y=-£ .
4. 7y + 8z=-31, a i« + i^ = 6,
12s-3y=-3. ^ag + ty=V-
5. 4a; + 5y=29, 9. 28a-45y = -17,
6a? + 7y = 41 . 35a?-27y = 8.
6. £a-£y=l, 10. 36® + 25y = ll,
^a? + ^y = 5 . 24 a? + 55y= -31 .
47. Second Method. Substitution.
Ex. Solve 2x+ y = % (1)
3a-2y = 10 . (2)
Solve (1) for y, y = 9 - 2 x. (3)
Show that this is transposition.
Substitute the value of y in (3) in (2).
3z-2(9-2x) = 10. (4)
Then, 3x-18 + 4x = 10
(How do you account for the signs of 18 and 4 a?)
Transposing and collecting, 7 x = 28. „
x = 4. (5)
Substituting (5) in (3), y = 9 - 2 • 4,
or, y = 1.
Check: 2-4 + 1=9.
EXERCISE 16
Solve by method of substitution (verify results) :
1. 4a>+ y = 6, 4. 42Z-6/S'=3,
3a? + 4y = ll . 82?4-3ff = l.
2. 12m- 7u=-2, 5. 15y - 8z = 11,
llm-12^= -13 . 35y-h6z=-32.
3. 422- 6£ = -1, 6. 10*- 8w = 7,
8JB-f3„S' = 3. fct-Y\^u=-— V
38 MATHEMATICS
48. Third Method. Comparison.
Solve each equation for the same variable and equate the
values thus found-
Ex. 1. 2x+ y = 9 y (1)
3s-2y=10. (2)
Solve (1) for y, y = 9 - 2 X. (3)
Solve (2) for y, y = 3 x ~ 10 . (4)
Equate the values of y found in (3) and (4) (Ax. 8).
9_2g = 8x - 10 - (5)
2 ,
3x-l(V
Multiply (5) by 2, 2^9 - 2 a = 3x ~~ 10 )
or, 18-4x = 3x-10. (6)
Solving (6) , a; = 4.
Substitute in (3) , y = 9 - 8
= 1.
Check as before.
Ex.2. 8x-5y=31, (1)
12x + 13y= -15 . (2)
Solve (1) for x, 8 x = 5 y + 31,
• = 8 -*±S. (3)
Solve (2) for a;, 12 x = - 13 y — 16,
8= -i8r-i6. (4)
12 v '
(5)
Equate the values of x in (3) and (4) (Ax. 8) .
5y + 31_ -13y-15
8 12
Multiply (5) by 24, a multiple of the denominators (Ax. 3).
Then, 15 y + 93 = - 26 y - 30. (6)
(See § 46, and note that the terms of (6) are the same as those that
compose equation (5), example 2, before they were combined into two
terms.)
INEQUALITIES 39
Transposing the unknown term to the first member, the known to
the second member, and solving,
41 y = - 123,
*=-S. (7)
Substitute in (3), s= ~ 15 + 31
8
= 2.
Check as before.
5.
6x
+ 3y =
3
~2'
5x
-3y =
:4.
6.
Sx
+ 3y
= 2,
9x
-15y
^ —
2.
EXERCISE 17
Solve by comparison, verifying each result:
1. 3x + 6y=7, 4. 15a-17.y = 16,
4a-2y = l . 13s4-Hy = l.
2. 5 y + 4 a; = 6,
2y- a? = l.
3. 8 x + 15 y = — 8,
43y-12a?==12.
49. In stating written problems it is often convenient to
use two or more unknown numbers rather than form a single
equation in one unknown as was done in exercise 8. Care
must be taken to form as many equations as there are un-
known numbers.
Ex. A number is composed of two digits. One half the
number is equal to twice the sum of the digits, and if 18 is
added to the number the digits of the resulting number will be
those of the original number written in reverse order. Find
the number.
Let t = the digit in tens' place,
and u — the digit in units 1 place.
Then, 10t + « = the number. (Exercise 2, example 14.)
By the conditions, 10t + M = 2 (t + u\ (1)
10* + w + 18 = 10tt + t,
or 9*-9m=-18,
40 MATHEMATICS
From (1) 6$-3m = 0,
or 2t- w = 0. (3)
Subtracting (2) from (3) t = 2. (4)
Substituting the value of t found in (4) in (2)
2 - u = - 2,
and u = 4.
Whence, 10 £ + w = 24, the required number.
EXERCISE 18
1. Find two numbers such that the sum of 3 times the first
and 5 times the second is 24, and the sum of £ of the first and
\ of the second is 2.
2. The altitude of a trapezoid is '8' and its area 48 square
feet. If the lower base is increased 4', its area will be increased
16 square feet. Find its upper and lower bases. (Any trouble
here ?)
3. A cubic foot of steel and a cubic foot of water together
weigh 552% poundg. The difference between their weights is
427^ pounds. How many times heavier than water is steel ?
This quotient is called the specific gravity of steel. (Remem-
ber these results.)
4. The perimeter of a rectangular field is 240 rods, and its
length is 40 rods more than its breadth. Find its area.
5. A field of wheat is 80 rods longer than it is wide. The
farmer uses a combined harvester and thresher that cuts an 11-
foot swath. After making 15 rounds, the indicator shows that
27£ acres have been cut and have yielded 660 bushels. Find
the yield per acre and the dimensions of the field.
6. Find two numbers such that n times the first added to k
times the second is c, and k times the first minus n times the
second is b.
7. The difference between the bases of a trapezoid is 6, the
area 120, and the sum of the bases 30. Find the dimensions.
Can you construct the trapezoid?
INEQUALITIES 41
50. When a problem involves three variables the method of
solution is similar to that in two variables. Care must be
taken to eliminate the same unknown in each equation.
In solving simultaneous equations it is necessary that there
be as many equations as there are variables in the problem.
Ex. 2x-y + 5z = 15, (1)
Sx + 2y -3z = -2, (2)
4rx-2y + 7z = 21. (3)
Multiply (1) by 2 and add the result to (2),
7ac + 7* = 28. (4)
Add (2) to (3), 7 x + 4 g = 19. (5)
We now have two equations in x and z.
Subtract (5) from (4), 33 = 9,
z = S. (6)
Substitute value of z in (5) ,
7 x + 12 = 19,
or x = 1. (7)
Substitute (7) and (6) in (1),
2-y + 16 = 16, (1)
or y = 2.
Check as before, substituting these values in (2) and (3).
EXERCISE 19
Solve the following :
1. 2s + 3y-4z = 12,
x—2y + 5z= —6,
3s + 2y-6z = 13.
2. 5y — 3x + 2z = 5,
5z-3y + 2x=-9,
5a-3z4-2,y = 12.
3. 3x—2y = 5,
3y-2z = 25,
3z-2x=-25.
4.
1 1 ,3„ 25
2 4 J 2 4'
1,2 3 A
5 5 o
2 2,1 5
-x y 4- -z = •
3 3* 3 3
5.
1 . 1 L l 13
-x+ -y-\--z — — ,
4 3* 2 12'
1,1 . 1 13
3 2 B 4 12'
-5C4--ti-V-Z =
42 MATHEMATICS
6. An examination contained 15 problems. Each pupil
received 7 credits for a problem solved, and 3 demerits for a
problem missed. One boy had 45 marks more to his credit
than he had against him. How many problems did he miss?
7. Find 3 numbers such that if 3 times the first be added to
one half the sum of the second and third, the sum will be 137.
If half the sum of the second and third be subtracted from the
first the difference will be 23, the sum of the three numbers
being 74.
8. If the order of the digits of a certain three-place number
be inverted the sum of this number and the original number is
444, and their difference is 198. The first digit is equal to the
sum of the other two. Find the number.
9. A merchant buys lemons, some at 4 for 5 cents, others at
3 for 4 cents, spending $ 2.80. Again he buys as many of the
first kind as he had bought of the second kind, and as many of
the second kind as he had bought of the first, this time spend-
ing $ 2.78. How many of each kind were purchased ?
Solution of Inequalities
51. In solving inequalities the rules for solving equations
hold except in some instances where change of sign takes
place (§§37-41).
EXERCISE 20
If x be positive, which fraction has the greater value ?
, x + 5 05 + 7 n 2a?4-9 2a?4-5
1. -J — or — — • 2. — J — or — -^ —
4 6 .62
(Test 1 and 2 by substituting values for x.)
Find the limits of x for which these are true, and show that
your results are correct:
3. Sx + B<x + 2* 5 - 6 *~ 4 < '
- 7cc4-14^ A
4. 5z-2>2a>4-7. 6 - — 9 >a
* Solve as you would an eqxiaWon.
INEQUALITIES 43
Find the limits of the values of x which satisfy the following
inequalities simultaneously (verify your results) :
7. 2z + 6>0. (1)
-a + 5>0. (2)
From (1), x>-3: (3)
From (2), z-6<0 (§40),
and x < 6.
Then, to satisfy the conditions of (1) and (2), x must lie between — 3
and + 5.
8. 24-3«>0, 10. 4z-7 <0,
16+2a;>0. 6x-2x>0.
9. 5a + 10>0, 11. 5z + 2 >0,
3a?- 9<0. 2j?-5a?>0.
What values of x and y satisfy the following (§ 41) :
12. 2a + 3y>10,
3s-2y = 8.
13. Seven times the number of boys in the algebra class less
45 is greater than five times their number added to 13, and
six times the number of boys less 22 is less than four times
the number of boys added to 40. How many boys in the
class ?
14. Find the digit such that twice the digit increased by 3
is greater than one fifth the digit increased by 11.
15. Find a multiple of 18 such that five times the number
decreased by 220 is greater than three times the number de-
creased by 42. Is there more than one such multiple ?
16. Find a multiple of 16 such that one half of it decreased
by 28 is less than one third of it increased by 16. Does more
than one multiple satisfy this condition ?
17. Find a line containing an integral number of inches
such that five times the length of it decreased by 165 is
greater than three times its length increased by 14. Is there
more than one such line ?
44 MATHEMATICS
1. The 20th Century Limited of the Lake Shore Railroad
runs 54.5 miles per hoar. Find the speed in feet per second.
2. There are 24 gas stores in the domestic science kitchen.
If each stove burns 2.32 cubic inches of gas per second, find
the cost of gas for one class of pupils daring a period of 80
minutes. Allow 15 minutes waste time (when stoves are not
in use). (Gas at 75^ per thousand cubic feet.)
3. The power of running a motor in the pattern shop is .044
Kilowatt per hour. During a period of 80 minutes, 24 motors
are running on an average of 60 minutes. Find the amount of
power used. Find the cost at $0.0635 per Kilowatt hour.
4. The Southern Talc Company is able to ship freight in car-
load lots, by the long ton, and charge its customers the cost price
per ton F. O. B. North Carolina, and base its charges on the
short ton. Find the Talc Company's profit on 10 cars of 34
tons each, shipped from North Carolina to Chicago. Freight
rate $ 3.50 per ton.
5. To find the weight of timber in the rough: multiply
length in feet by breadth in feet by thickness in inches and
multiply that product by one of the following factors. For
Oak, 4.04 ; Elm, 3.05 ; Yellow Pine, 3.44 ; White Fine, 2.97.
The result is in pounds. Measure six pieces of lumber in the
shops and find the weight of each piece. Make accurate draw-
ings of each piece.
6. The weight of a grindstone is found by multiplying the
square of the diameter in inches by thickness in inches and
this product by .06363.
(Diameter) 2 (Xhickness) (.06363) = weight in pounds. (Re-
member this formula.) Find weight of grindstone in your
shop. Make diagram.
7. The weight of one cubic foot of water is 62£ pounds.
Find the weight of one gallon.
INEQUALITIES 45
8. Stock must be ordered for making 10,000 steel pins, each
%" diameter by 2" long. The rods for making these pins are \ n
in diameter and 12-0" long. Allow ^ n waste in cutting each
pin. How many rods will be needed ? What will they weigh ?
9. For flavoring, 1 ounce of chocolate is equivalent to 2 table-
spoonfuls of cocoa. Chocolate costs 22 cents a cake (8 ounces),
while cocoa costs 25 cents per box (£ pound). There are 2 cups
of cocoa in one box. Which is cheaper to use ?
10. Chocolate contains 48.7% fat, and cocoa, 28.9%.. If
cocoa were used in place of 2 ounces of chocolate, how many
tablespoohfuls of butter would need to be added to make the
same quantity of fat as in chocolate, butter containing 85'%
fat, and one tablespoonf ul weighing \ ounce ?
11. Butter contains 85% fat, cream 18.5% fat, milk 4% fat.
Calculate the quantity of butter necessary to use with 2 cups
of milk to produce the same quantity of fat as 2 tablespoon-
f uls of butter and 2 cups of cream, one pint of cream weighing
one pound, and one pint of milk one pound and one ounce.
12. Potatoes pared and then boiled lose 2.7% starch; pota-
toes boiled with the skins on lose .2% starch. Find difference
in quantity of starch lost by cooking 6 potatoes, one potato
weighing 4£ ounces.
13. Boiled Irish potatoes contain 20.9 % carbohydrates.
Cooked sweet potatoes contain 42% carbohydrates. An Irish
potato weighs 4 J ounces, and a sweet potato 6 ounces. Find
difference in quantity of carbohydrates in six of each kind of
potatoes.
14. Find the reciprocal of 5280. Would you rather divide a
number by 5280 or multiply the same number by its reciprocal ?
Which would give the larger result ? Why ?
15. Multiply 18649 cubic inches by the reciprocal of 1728.
Is the result cubic feet ?
16. Find the weight of a grindstone 36 " in diameter and 4^ ,(
thick. What will it weigh after it has bsfcm. ^ora. &.orcra.1\ x '\
46 MATHEMATICS
17. A class of 15 girls wish materials and thread for cooking
aprons. Each apron requires 2\ yards muslin at 12^ per yard,
and a spool of thread at 5^. How much material and thread
for the class ?
18. How much money was spent in making the purchases in
example 17 ? How much does each girl pay for her share ?
19. If a girl earned 22 \ $ a piece making these aprons, how
many would she have to make to earn $ 5.50 a week ?
20. If these aprons sold at 85^ each, what would be the
profit per apron ?
)
CHAPTER V
Lines, Angles, Triangles
52. A portion of space is called a geometric solid. The
boundary between the solid and the remaining space is called
the surface of the solid. For example, if the sides, bottom, and
top of a box were considered to have
no thickness, they might be called the
surface of the solid.
53. If two surfaces intersect, their
intersection is called a line. Thus, if
^^ jf»^2) surfaces AB and DO intersect in XCO y their in-
tersection, XCO, is called a line.
54. An intersection of two lines is called a
point. Thus, F, the intersection of lines AB
and CD, is a point.
55. A line may also be regarded as ^/^ ^> — , ^y
the path of a moving point.
56. The line is said to be straight when it does not change
its direction at any point.
A line has no limit in its length. When a part of it is in-
dicated, we speak of it as a segment or sect of the line. Thus,
A By AB indicates a straight line passing
through points A and B. The part of the line between A and
B is a segment or sect of the line. A line may also be indi-
cated by a small letter placed somewhere on the line.
E.g., — is read " the \me a"
47
A
48 MATHEMATICS
B 57. A curved fine changes its direc-
tion ac every point. £^- y the curve
C ABC.
58. A brakes fine is composed of segments of snccessive
p, -p straight lines which have different
J 7 directions and which, pair by pair,
O D G have a point in common. E.g n
ABCDEFG is a broken line.
59. The word line, unmodified, usually means a straight
line.
60. A plane is a surface such that if any two of its points
Mi i be joined by a straight line, the
line will lie entirely in the sur-
face.
Thus, in plane MX, if any two
points, A and B, are joined by a
line, the line will lie entirely within the surface.
61. Plane Geometry treats of combinations of points and
lines all lying in the same plane.
62. A circle is a portion of a plane bounded by a curved
line every point of which is equidis-
tant from a point within called the cen-
ter.
The curved line is the circumference.
The distance from the center to any
point in the circumference is the radius,
as CD.
An arc is any part of the circumference, as AB.
LINES, ANGLES, TRIANGLES 49
Problem
63. To draw a straight line equal to a given straight line.
• B
A
Given a line MN.
To draw a straight line equal to MN.
1. Draw an indefinite line AB.
2. With A as a center and MN as a radius, describe an arc
cutting AB at D.
3. AD is the required line.
64. An angle is the figure formed by two lines drawn from a
point.
The point is called the vertex.
The above angle is read ABC, the letter at the vertex always
£ being between the other two. The angle may
also be designated by a small letter placed
between the sides of the angle, as the angle x.
This may be written Zx* The size of an
angle does not depend upon the length of the
sides, but upon the difference in the direction of the sides.
65. Two angles are said to be adjacent when they
have a common vertex and a common side between
them.
Thus Z a and Z b are adjacent.
66. Two angles are equal when one can be so applied to the
other that they will coincide throughout.
* Also when only one angle is formed at a vertex, it may be read by a single
letter, as Z. B, read " angle at B." Small letters denote ualue%\ toppVtaXVxvKr*
denote position only.
50
MATHEMATICS
Problem
67. lo construct an angle equal to a given angle.
Given A ABC.
To construct an angle equal to it.
1. Draw any line MN.
2. Take any point D on MN for
the vertex of the angle.
3. With B as center and any
radius BL, describe an arc cutting A B
and BC at L and K, respectively.
.4. With BL as radius and D as
center, describe an arc cutting DN at
O.
5. With as center and distance
LK as radius, describe an arc cutting
the former arc at E.
6. Draw DEF.
7. Then NDF is the required angle.
68. A theorem is a truth requiring proof.
In proving theorems the following axioms, in addition to
those in § 23, are useful :
Ax. 9. But one straight line can be drawn between tivo points.
Ax. 10. A straight line is the shortest distance between two
points.
69. A triangle is a portion of a plane bounded by three
straight lines. The lines are called the sides of the triangle.
The points where the sides meet are the vertices.
Three-sided figures play an important part in the study of
geometry and in its applications to many problems in mathe-
matics. Much of the work in measurement, in surveying, and
'n other kinds of engineering depends upon, the triangle.
to
^s
\ ^
LINES, ANGLES, TRIANGLES
51
Theorem I
70. Two triangles are equal when two sides and the included
angle of one are equal, respectively , to two sides and the included
angle of the other.
Draw a A ABC*
On any line DE take DN equal
to AB (§ 63).
At D construct ZD = /.A
(§ 67).
On Intake DM = AC and draw
MN.
In the above theorem, we have
three equations given. (This given j)dL
part is called the hypothesis.)
sF
M/
*r-
?s
\.
'* i
w
-E
The given equations are :
AC = DM
AB = DN
ZA = ZD
(Don't forget that what is
given, the parts that are given,
are the tools you have to work
with.)
To prove the geometric equation,
A ABC = A DNM. (This is called the conclusion.)
Proof. 1. Place A DNM on A ABC in such a manner that
DN will coincide with its equal AB, D falling on A.
2. DM will take the direction AC because Z D = Z A.
3. Point M will fall on point C since DM= AC.
4. Hence line MN will coincide with line CB (Ax. 9).
5. Then A ABC = A DNM, since they coincide throughout.
* In reading the letters at the vertices of a triangle always read counter-
clockwise, beginning as far as is practicable at the lower left-hand CAT^feii.^.^^
ABC, DNM in above triangles.
52 MATHEMATICS
Theorem II
71. Two triangles are equal when a side and two adjacent
angles of the one are equal, respectively, to a side and two adjacent
angles of the other.
/ \
9 *
/ \
\
X
/ \ / x
/ \ \
d< m *JT
What is the hypothesis ? What is the conclusion ?
Draw a A u4BC.
Draw a line ZW equal to AB.
At Z> construct an angle equal to Z.A.
At ^T construct an angle equal to Z B.
Extend the sides of these angles to meet at M. We have
Given in the A ABC and DNM
DN=AB,
ZD = ZA,
ZN=ZB.
To prove A ABC = A DNM.
Proof. 1. Place A DNM upon A ABC in such a way that
2)^T will coincide with its equal AB, point D falling on A,
2. Since AD = /.A, side DJf will take the direction AC
and Jlf will fall somewhere on AC
3. Since ZN=ZB, side JVJIf will take the direction BC and
M will fall somewhere on BC
4. Since Jlf falls on both AC and 22(7, it must fall at C, their
point of intersection.
5. Hence A ABC and DNM coincide throughout and are
equal
LINES, ANGLES, TRIANGLES
53
Problem
72. To bisect a given angle.
1. Draw an angle DCE.
2. With C as center and any radius CM, describe an arc
cutting CE and CD at M and JV", respectively.
3. With M and N as centers and a radius greater than one
half MN, describe arcs intersecting at 0.
4. Draw CO.
5. Then CO is the bisector of Z DCE.
73. The right angle. If one straight line meets another
straight line in such a manner that the adjacent angles (§ 65)
formed are equal, the angles are called right angles and the
lines are perpendicular to each other.
Thus, if AB is a straight line and Zaj = Zy, x and y are
each right angles, and MP is perpen-
dicular to AB.
( J_ is the abbreviation for " perpen- .
dicular to.")
x
V
M
B
74. If one straight line meets another straight line, adjacent
angles are formed. If these angles are not right angles, the
one greater than a right angle is called obtuse. The one less
than a right angle is called acute. Either angle is said to be
oblique.
54
MATHEMATICS
D
75. If a line CD meets another line AB, two adjacent angles
x and y, are formed. Suppose y less than x. If CD is made
to rotate toward A, about D as a
pivot, Z a; diminishes and Z y in-
creases. It is evident that at
B some position DK of DC, Z <c will
equal Z.y. There can be only
one such position. Then, one and only one perpendicular can
be erected to a given line at a given point in the line.
76. It follows from § 75 that :
(a) Two right angles are equal.
(b) If one straight line meets another straight line, the sum oj
the adjacent angles formed is equal to two right angles. That is,
in the figure of § 75.
Z x + Z y = Z ADK + Z KDB = 2 rt. A.
(c) TJie sum of all the angles on the same side of a straight line
at a given point is equal to two right angles.
(d) The sum of all the angles about any
point in a plane is equal to four right angles.
a + b +c + d+ e +/=4 rt. A.
(e) If the sum of two adjacent angles is
two right angles, their exterior sides lie in
the same straight line.
Let a -f b = 2 rt. A.
Extend PM to E. If AM and MB
were not in a straight line, the sum Ar
of c and d would not be two right
angles, and this is contrary to (d).
/
'E
9Md
-B
77. Species of triangles.
The right triangle is a triangle one of whose angles; is 90 c
LINES, ANGLES, TRIANGLES
55
or a right angle. The side opposite the right angle is
called the hypotenuse. The other two sides are called the
legs.
All triangles except right triangles are oblique triangles.
If an oblique triangle contains one obtuse angle, it is called
an obtuse angled triangle.
An acute angled triangle has each angle less than a right
angle.
Triangles are also named according to their lengths of
sides.
A scalene triangle has no two sides equal. An isosceles
triangle has two sides equal. An equilateral triangle has three
sides equal.
~JB
Theorem III
78. In an isosceles triangle the angles opposite the equal sides
are equal.
Draw line AB. With A and B as
centers and the same radius greater
than \ AB, describe arcs intersecting
at P. Draw PA and PB. We now
have the isosceles triangle ABP, with
b = a.
What is given ?
To prove A B = A A.
Proof. 1. Draw line PM bisecting
APB (§, 72).
2. In A AMP and MBP,
b= a; h = h; Az = Ay.
3. Hence, A AMP = A MBP
(§ 70).
4. Then AB = A A, being like parts of ex\\ial f\%\sc«&.
56 MATHEMATICS
79. In equal figures, corresponding lines or angles are called
homologous. It follows that in equal figures homologous parts
are equal.
Note. In equal tr (angles equal sides lie opposite equal angles.
Thus, in equal A AMP and MBP, § 78, side a = side b, then
Z A is homologous to Z B.
Problem
80. To construct a triangle when three sides are given.
Let the given sides be a, b, c.
1. Draw an indefinite
line AK.
2. On AK take AB = e.
3. With A as center and
b as radius, describe an arc. \
4. With B as center and c
a as radius, describe an arc
intersecting with the former arc at C.
5. Draw AC and CB.
6. Then ABO is the required triangle.
Theorem IV
81. Two triangles are equal when three sides of one are equal,
respectively, to three sides of the other.
Draw a triangle ABC.
Construct a second triangle DEF whose sides are equal to
x, b, c, respectively (§ 80). Namely, d =a, e = b,f=c.
What is given ?
To prove A DEF = A ABC
Proof. 1. Apply A DEF to A ABC in such a way that /
will coincide with its equal c, and F will fall opposite C.
2. Draw CF.
LINES, ANGLES, TRIANGLES 57
3. In A AFC, b = e. (By Hypothesis.)
4. Then Zx = Zy (§ 78).
(In an isosceles triangle the angles opposite the equal sides are equal.)
e/
/ x
/
V
s
s
/
v
N
D* J h E
5. Similarly, A FBC is isosceles and Zn=Zm.
6. Adding the equations in steps 4 and 5,
7. Then, in A ABC and AFB,
b = e
Z ACB, or Zx+Zn=Z AFB or Zy + Zm.
8. Hence, A ACB = A AFB (§ 70).
(Two triangles are equal when two sides and the included angle of the
one are equal, respectively, to two sides and the included angle of the
other.)
9. But A AFB is the same as A DEF.
10. Hence, A DEF = A ABC
Note. The base of a triangle is the side on which it is sup-
posed to rest. Any side may be considered the base. The
altitude is a perpendicular (§ 73) from any vertex to the oppo-
site side. Theorems are truths to be used. Eemember each
one. Notice that theorems I and III are both \ise<L \sl ^earcva^
theorem TV.
58
MATHEMATICS
EXERCISE 21
1. Construct a triangle each of whose sides is 8".
2. The sides of a triangle are 3", 4", 5", respectively.
Construct the triangle.
3. The sides of a triangle are 13', 15', 16', respectively.
Draw this triangle to a scale, using |" = 1' — 0".
4. Draw three altitudes of the triangle constructed in
example 3.
5. Construct a triangle whose sides are 15', 16', 28'. Draw
the three altitudes.
6. Measure the three altitudes of the triangle constructed
in example 1. How do they compare? Is this comparison
the same in altitudes of other triangles of this exercise ?.
£ 7. In the figure ABCD, in & ABD
, z and CDB, x = y and o — z. Compare
the triangles. Give reasons for your
conclusion.
8. In the figure ABCD, in triangles
ABC and CDA, x is equal to y, o is
equal to z, Z D is equal to Z B. Com-
pare the triangles. Give reason for
your conclusion.
9. In triangles ABD and CDB,
Z.X — /.Z, and Zy = /.o. Compare the
triangles. Why ?
10. If in triangle A BO, b = a, and CD is C
so drawn that AD = DB, compare triangle
ADC with triangle CDB.
11. On a six-inch base construct a triangle
whose remaining sides are each 14". What do A
you know about this triangle ?
LINES, ANGLES, TRIANGLES 59
12. The sum of two numbers is 24. One number is four
more than the other. What are the numbers ? (See example
10, exercise 8.)
13. The difference of two numbers is 8, and their product
is 160 more than the square of the less number. Find the
numbers.
14. The sides of a right triangle (§ 77) are as 4 to 3.* The
square of the hypotenuse is equal to the sum of the squares of
the other two sides. The hypotenuse is 25. Find the sides
and the area.
15. In a lever AM, P= power; F= fulcrum, on which the
lever rests; W= weight. Note that if the
lever is pressed downward at P, W tends |*[ 5
to rise. It has been found that: power M2-(& 6 L 0' A
times the distance, from where the power is
applied, to the fulcrum is equal to the weight times the distance
from the weight to the fulcrum, or
P-AF=W-MF.
AF is called the power or force arm, MF, the weight arm.
If the power arm is 6' — 0", the weight arm 2' — 0", and
the weight 200 pounds, what power is necessary to lift the
weight ?
Note. In these problems, the weight of the lever is neglected.
16. With a 4" x 4", 14 feet long, used as a lever, a 100-
pound boy on the end of the power arm
is just able to lift a weight W, when
.the fulcrum is 2' — 0" from the weight.
What weight does he lift ?
* In such statements fractions are avoided by letting the required values be
represented by a multiple of the unknown number. For example* lAt.^Vs^.
one Bide, and 4s the other.
60
MATHEMATICS
T — --(
1 1
1
1 '
F5
«e- ,
1
1
i
1
17. A crowbar is 5' — 0" long. The
fulcrum is 3' f from one end. A force of
20 pounds at the other end of the bar is
required to pull a railroad spike from a tie.
With what force does the spike hold ?
18. With the crowbar and fulcrum as
in example 17, two boys weighing 100
lb. and 115 lb., respectively, were able >
to move a freight car standing on a sid- (Tl) (TJ
ing. What force was necessary to start \*$
the car?
19. The length and breadth of a rectangular tank are as 4 to
3. The water in the tank is frozen to a depth equal to one
fourth the width of the tank. The area of the bottom of the
tank is 48 square feet. Find the weight of the ice. (Ice
weighs 57£ pounds per cubic foot.) Which is heavier, water
or ice? Why is this necessary ? See exercise 18, example 3.
20. Five lines are drawn from a point forming
angles a, b, c, d, e. The sum of a and b is 90° ; d is
equal to 12° more than a; c is equal to 6; e is equal
to twice d. Find the angles. (See § 76, d.) Is the drawing
correct ?
21. Divide 48 into three parts, such that the first part shall
be twice the second, and the third 8 more than the first.
22. A rectangle and a square have the same altitude. The
base of the rectangle is 8' more than the base of the square.
The areas differ by 64 square feet. Find the dimensions of
each.
82. An acute angle is an angle that is less than a right
angle. Ex. Angle a.
-a An obtuse angle is an angle that is greater than a
• right angle and less than two right angles. Ex.
- Angle b.
LINES, ANGLES, TRIANGLES 61
83. When two straight lines intersect, the opposite angles
formed are called vertical angles.
Thus, a and c are vertical angles, also b and d.
84. An angle is measured by comparing it with
another angle considered as the unit of measure.
The most general unit of measure is -fa of a right
angle, and is called a degree.
The degree is divided into 60 equal parts called minutes,
and the minute into 60 equal parts called seconds.
The abbreviations for degrees, minutes, seconds, are °, ', ",
respectively.
85. The right angle (§ 73) is measured by one fourth of the
circumference of a circle, or 90°.
Q 86. If the sum of two angles is a right
angle, each is a complement of the other,
and the angles are said to be complementary.
Thus, if ABC is 90°, a is the complement
^ u, j of b, their sum being 90°.
87. If the sum of two angles is 180° or two right angles,
each is the supplement of the other, and the angles are said to
be supplementary.
Thus, if the sum of angles c and d is 180°, c
d/ / and d are supplementary and each is the
supplement of the other.
88. It follows from § 87, that if two supplementary angles,
as c and d in the figure are adjacent (§ 65), their exterior sides
lie in the same straight line.
Then c and d are supplementary adjacent angles.
EXERCISE 22
1. Find the complement of 38°; the supplement of 38°.
2. Are 36° 30' and 53° 30' comptamertwrg ** ^Vj 5 *
62 MATHEMATICS
3. Are 130° 19" and 49° 41' supplementary ? Why ?
4. Find the complement of each of these angles : 24° ; 36° ;
72°; 60°; 30°; 45°; 30° 15'; 24° 15' 20"; 47° 10.5'; b°.
5. Find the supplement of each of these angles : 22° ; 96° ;
120°; 150°; 24° 8'; 27° 19' 36"; 45° 4.4'; 90°; 6°.
6. Find the complement of the supplement of : 120° ; 130° ;
135° ; c°. Does this last result give a simple formula ?
7. Find the supplement of the complement of: 3°; 24°;
28° 5' 30"; 45°; 90°; c°. Does this last result afford a means
of simplifying other parts of this example ? Illustrate.
8. Angle at A = 36°, ZB = 74°. Find the supplement of
their sum.
9. Angle at A = 29°, Z B = 31°. Find the complement of
their sum.
10. Angle at A = 75°, Z. B = 15°. Find the complement of
their difference.
Theorem V
89. Any side of a triangle is greater than the difference of the
other two sides.
Draw any triangle ABC, the sides opposite A, B> C 9 being
a, by c, respectively.
We have
Given a any side of A ABC, and c> 6.
To prove a>c—b.
Proof. 1. a + b>c.
A straight line is the shortest distance between two points.
(§ 68, Ax. 10).
2. a>c — b. (Transposing b, § 42.)
3. Hence, any side, a, is greater than the difference of the
other two sides.
LINES, ANGLES, TRIANGLES 63
Theorem VI
90. The sum of two sides of a triangle is greater than the sum
of two lines drawn from any point within the triangle to the
extremities of the third side of the triangle.
Draw* any triangle ABC.
From any point P, within the triangle draw PB and PC.
Call side AB, c ; AC, b ; PB, e ; PC, d.
We now have:
Given A ABC, with lines d and e drawn from any point P
within the triangle.
To prove c -f- b > d + e.
Proof. 1. Produce BP to E.j
Call PE, o ; AE, x ; EC, y.
2. c + x > e + o.
A straight line is the shortest distance between two points
(§ 68, Ax. 10).
3. o + y > d (§ 68, Ax. 10).
4. Add 2 and 3,
5. Subtract o from each side of this inequality, .
c+s+y>d+ e (§39).
6. But a; + y = 6 (§ 23, Ax. 6).
7. Whence, substituting for x+y its equal 6,
c + & > d + e.
* This description of the drawing is not a part of the demonstration. Your
demonstration depends solely on the three paragraphs: Given, To prove,
Proof. Be sure you know what you have given you to work with and what
you wish to prove.
t Remember that any additional lines you draw Va. a. ^px* \br&\.\a fcaNfrak.
lines.
64 MATHEMATICS
Theorem VII
91. If two straight lines intersect, the vertical angles are equal.
Draw two intersecting I"
lines, MN and X Y, form-
ing the angles a, b, c, d. ^ ^yf
We then have
Given two intersecting
lines, MN and XY, and
vertical angles a, c, and b, d.
To prove a = c and b = d.
Proof. 1. a + b = 2 rt. A (§ 88).
2. & + c = 2 rt. .4 (§ 88).
3. Subtract 2 from 1 (§ 23, Ax. 2),
a — c = 0,
or a = c.
4. The pupil may prove b = d.
What is the hypothesis in the above theorem? What is
the conclusion ?
Problem
92. To draw a perpendicular bisector of a straight line.
Draw line AB.
To draw a perpendicular bisector of AB,
1. With A as center and any radius greater than one half
AB, describe arcs on each side of line AB.
2. With B as center and the same radius, describe arcs in-
tersecting the arcs already drawn at C and at D.
3. Draw CD.
4. Then CD is the perpendicular bisector of AB.
The reason for this construction will be given in § 95.
F, the point of intersection of CD and AB, is called the foot of
the perpendicular.
Note that O and D are each. equally &\&taa\> itom A ^eA. Tk»
LINES, ANGLES, TRIANGLES 65
Problem
93. To draw a perpendicular to a line at any point in
the line.
1. Draw a line AB.
2. Take any point P in the line AB.
3. To erect a perpendicular to line AB at P.
4. With P as center and any radius, describe arcs inter-
secting AB, or AB produced, at C and E.
5. With C and E as centers and radius greater than one
half CE, describe arcs intersecting on one side of AB.
6. The pupil may show that the perpendicular may now
be drawn and that this problem is an application of § 92.
Why must the radius be greater than one half CE ?
EXERCISE 23
1. How many points determine a line ? What do you mean
by determine as used in this sense ?
2. How many conditions can be imposed on a line ? What
properties besides length has a straight line ?
Theorem VIII
94. If a perpendicular is erected at the middle point of a line,
I, Any point in the perpendicular is equidistant from the ex-
tremities of the line.
II. Any point not in the perpendicular is unequally distant
from the extremities of the line.
Draw line KF± line AB at its middle point F.
From P, any point in KF, draw lines PA and PB.
We then have,
I. Given two lines PA (x) and PB (y) drawn from any
point P in the perpendicular to AB at its middle point.
To prove x = y.
66
MATHEMATICS
Proof. 1. In triangles AFP and PFB,
o = b (Constr.)
h = h (Identical)
ZAFP=Z.BFP.
(All rt. A are equal.)
2. Then, A AFP = A PFB.
(Two triangles are equal when two sides, etc., § 70.)
3. Then, x = y.
(In equal figures corresponding parts are equal.)
Does this remind you of Theorem III ?
II. Given P ' any point not in KF.
To prove PA =£ P*B*
or x + c =£ m.
Proof. 1. If P' is not in KF, P'A or
P'B must intersect KF.
Suppose P 'A intersects KF at P. j±
Draw PB (y).
2. Then, y + c =£ m (§68, Ax. 10).
3. But y = x (§94,1).
4. Then, substituting x for its equal y in 2,
a? + c =£ m.
95. From the result of § 94, we may conclude that :
I. A point equally distant from the extremities of a line lies
in the perpendicular at the middle point of the line.
II. Two points each equally distant from the extremities of a
line determine the perpendicular at the middle point of the line.
III. Two lines drawn from a point in the perpendicular to
a line and cutting off equal distances from the foot of the per-
pendicular make equal angles with the perpendicular, and are
egual. (The proof of III is left to the pupil.)
* 4=. is read " does not equaX."
LINES, ANGLES, TRIANGLES 67
Problem
96. To draw a perpendicular to a line from a given point
without the line.
Given line AB and point P without AB.
To draw PF ± AB. Construction :
1. With P as center and a radius greater than the distance
from P to line AB, describe an arc intersecting AB at E and D.
2. With E and D as centers and the same radius greater
than one half ED, describe arcs intersecting at K.
3. Draw PK.
4. PKis the required perpendicular.
Note that P and K are each equally distant from E and D.
Compare EF and FD (§ 94).
Theorem IX
97. From a point without a line but one perpendicular can be
drawn to the line.
Draw line AB and PF1.AB, also PH any other line meet-
ing AB at H. We have
Given PF± AB, and PH any other line from P to AB.
To prove PH not perpendicular to AB.
Proof. 1. Produce PF to P' making FF = PF.
2. Draw HP'.
3. Represent A PHF and PHF by x and y, respectively.
4. Then, Z. x = Z y (§ 95, III).
5. PFP' is a straight line (by construction).
6. Then, PHP 1 is not a straight line.
(But one straight line can be drawn between two points, Ax. 9.)
7. Then, x + y =£ 180. (§88)
8. Whence, 2 x =£ 180, and a =£ 90.
9. Hence, PH is not perpendicular to AB. And since PH
is any line except PF, PF is the only per\re\id\c.>i\ax. l\ss& ^»x^
be drawn from P to line AB.
68
MATHEMATICS
EXERCISE 24
1. The base of an isosceles triangle is 8", the altitude 6".
Construct the triangle.
2. The altitude of an isosceles triangle is 8", one leg is 10".
Construct the triangle.
3. The sides of a triangle are 5", 12", 13", respectively.
Construct the triangle.
4. The base of a triangle is 13", the base angles each 45°.
Construct the triangle. Measure the angle at the vertex.*
5. The sides of a triangle are 4", 10", 6", respectively.
Construct the triangle. Explain your result.
6. Two sides and the included angle of a triangle are 8",
12", 45°, respectively. Construct the triangle.
7. Draw two acute angles, a and b. At a given point on a
line AB, construct an angle equal to the sum of these two
angles.
8. Draw a triangle ABC, and at a given point P, on a line
DE, construct an angle equal to the sum of the angles at A,
B, and C.
Theorem X
98. Two right triangles are equal when the hypotenuse and an
acute angle of the one are equal, respectively, to the hypotenuse
and acute angle of the other.
Draw right A ABC, and side A'B' (c') = side AB (c).
* The vertex opposite the base is the vertex oi ti& totaa^a.
LINES, ANGLES, TRIANGLES
69
At B' construct ZB' = ZB, and draw A'C _L BE. Repre-
sent BC, B'C by a and a', respectively, also AC, A'C by b
and b', respectively.
We then have
Given rt. A ABC and A' B'C with c = c', and Z B = Z £'.
To prove A ^BC = A A B'C.
Proof. 1. Place A ABC on A A 1 B'C in such a manner that
c will coincide with its equal c'.
2. Since Z £ = Z £', a will fall on a 1 .
3. -4 falls on A', then side b will coincide with side b'.
(But one X can be drawn from a point to a line from a
point without the line, § 97.)
4. Hence, the triangles are equal, since they coincide
throughout.
Theorem XI
99. Two right Mangles are equal when the hypotenuse and a
leg of the one are equal, re-
spectively, to the hypotenuse ■*>
and a leg of the other.
Draw rt. A ABC On
line DH, at G, erect EG _L
DH and equal to b. With
E as center and.c as radius,
describe an arc intersect-
ing DH at F. We then
have
Given two right trian-
gles, ABC and EFG, with
c = g, and b = /.
To prove A ABC = A EFG.
Proof. 1. Apply A EFG to A ABC in such a manner that
/will coincide with its equal b, vertex F ia\\Vii^ aX» "F?
70
MATHEMATICS
2. Since A ACB and ACF are right angles, BCF* is a
straight line (§ 76).
3. Since c = g A .B^' is isosceles, and Z £ = Z 2?" (§ 78).
4. Then, A ABC = A ACF' (§ 98).
5. Then, A ABC = A EFG. Why?
Theorem XII
100. If two unequal oblique lines drawn from a point in a
perpendicular to a line, cut off unequal distances from the foot
of the perpendicular y the more remote is the greater.
Draw line AB, and PF± AB at F, also lines PD and PC,
DF being greater than CF.
To prove PD > PC.
Extend PF to P, making FP' = PF.
Draw DP 1 and CP'.
The proof is left to the student ($ W).
LINES, ANGLES, TRIANGLES 71
Theorem XIII
101. If oblique lines are drawn from a point to a straight line
and a perpendicular is drawn from the point to the line,
I. Two equal oblique lines cut off equal distances from the
foot of the perpendicular.
II. The greater of two unequal oblique lines cuts off the greater
distance from the foot of the perpendicular.
1. Call AB the given line, P the given point, PF the per-
pendicular to AB, PE and PC the equal oblique lines.
The proof is left to the student.
(Prove A PFE = A PFC.)
II. Draw PF± AB, also lines PC (x) and PD (y), with x > y.
We then have
Given PF1. AB, and line x > line y.
To prove CF > DF.
Proof. 1. If CF= DF, x = y (§ 95, III). But this is con-
trary to hypothesis.
2. If CF<DF, x<y (§ 100). This is also contrary to
hypothesis.
3. Hence, if CF is not equal to or less than DF, it must be
greater than DF.
. 102. The manner of proving § 101 is called "The Indirect
Method." In §101 what is the hypothesis? What the
conclusion ? Compare § 95, III. What is the relation be-
tween the hypothesis of § 95, III, and the conclusion of § 101,
I ? When the conclusion of one theorem is the hypothesis of
the other, and the hypothesis of the one is the conclusion of
the other, each theorem is the converse of the other.
EXERCISE 25
1. The line joining the vertex of an isosceles triangle to the
middle point of the base bisects ttia N^xWsaX ^s^fe»
72 MATHEMATICS
2. Two isosceles triangles have a common base but unequal
altitudes. Show that the line connecting their vertices is per-
pendicular to the base and bisects the base.
3. If the altitude of a triangle bisects the base, the triangle
is isosceles.
4. Two isosceles triangles are equal if a leg and the vertical
angle of one are equal, respectively, to a leg and the vertical
angle of the other.
5. The hypotenuse of each of two right triangles is 81 ' — 6".
Each has an angle of 31° 41'. Compare the triangles.
6. The hypotenuse and one leg of a triangle are 3 yards,
1 foot, and 1 yard, 2 feet, and 6 inches, respectively. The
hypotenuse and one leg of another triangle are 10' — 0" and
5' — 6", respectively. Compare the triangles.
ORAL REVIEW
Give the results of the following :
1. 19-17. 10. 6.11-7.11.
2. 16-15. 11. 9.15 + 6.15.
3. 18-16. 12. 8.17 + 7.17.
4. 18-5-6.17. 13. 8 -16-1-8.16.
5. 24- 6 + 3- 17. 14. 27.3-4.28.
6. 4-18-3-16. 15. 10.34-5-17.
7. 24.5 + 3-17. 16. 12-8+12.8.
8. 9.8 + 5-7. 17. -3.15-3.15.
9. 6.15-3-14. 18. -8.16 + 7-15.
Supplemental Applied Mathematics
1. A grown person needs 3000 cu. ft. of air per hour that
the functions of the body may be active. If a room 20 ft. by
LINES/ ANGLES, TRIANGLES 73
15 ft. by 10 ft. were occupied by one person, how often would
the air have to be completely changed to obtain pure air ?
2. How often would the air in a room 12 ft. by 11 ft. by S\
ft. have to be completely changed to obtain pure air for two
persons ?
3. If there were 500 people in a lecture hall, how often
would the air have to be completely changed to insure good
ventilation ? Nine square feet of floor space is allowed to each
person, and the hall is 11 feet high.
4. For good ventilation the air in a room containing one'
person needs to be changed every 45 minutes. What is the
approximate size of the room ?
5. It has been found by experiment that the air in a room
cannot be changed more than three times per hour without
danger of drafts. What are the dimensions of a room that is
just large enough to meet this ventilating requirement, when
one person occupies the room ?
6. It has been found by experiment that one gas jet, when
burning, uses as much air as two persons. By changing the air
in a room occupied by one person once an hour during the
daytime, good ventilation is secured. How many times per
hour does it need to be changed in the evening when one gas
jet is burning ?
7. When two persons occupy a room, good ventilation is
secured in daytime by changing the air once each half hour.
How often should the air be changed when 3 gas jets are
lighted ?
8. The air in a room occupied by one person needs to be
changed once every hour in the daytime and three times every
hour during the evening. How many gas jets are burning?
9. A kerosene lamp requires as much air as 4 persons.
When 2 persons occupy a room, the air needs to be changed
once every 2 hours. When a lamp burns in the room, how
often does the air need to be changed.?
74 MATHEMATICS
10. In the daytime the air of a room containing 2 persons
needs to be changed once every 45 minutes, and in the evening
once every 15 minutes. The room is lighted by kerosene lamps.
How many kerosene lamps are burning in the room ?
11. The quantity of carbon dioxide given off by candles is
about twice as much as that given off by gas. If the air in a
room needs to be changed once every 1£ hours when illuminated
by gas, how often will it need to be changed when illuminated
by candles ?
12. For practical purposes architects figure 30 cubic feet of
air per minute for each person. A classroom has a ventilating
system. The room is 28' X 23' X 15' and contains 30 pupils.
To insure good ventilation, how much air must be driven into
the room and how many times per hour must the air be
changed ?
13. In hospitals it is customary to allow 50 cubic feet of air
per minute per person. In a hospital ward 56 f x 9' X 12' are 8
patients, a nurse, and 2 gas jets. How much air must be sup-
plied per hour? How many times must it be changed per
hour ?
14. A train running 40 miles an hour strikes two torpedoes
400 feet apart. Sound travels 1090 feet per second. What
time elapses between their reports at a station which the train
is nearing ?
15. According to some engineers, the sectional area of the
cold-air box of a furnace should be equal to the combined areas
of all the registers. There are 6 registers in a house, each 8 in.
by 10 in. How large should the cold-air box be ?
16. The cross section of a cold-air box is 2 ft. 3 in. by 1 ft.
8 in. There are 6 equal registers in the house. What is the
area of each register ?
17. The sectional area of a cold-air box is 495 sq. in. Each
register .measures 9 in. by 11 in. How many registers are
there in the house ?
LINES, ANGLES, TRIANGLES 75
18. According to other engineers, the sectional area of a cold-
air box should be equal to the combined areas of all the registers
minus one sixth. There are 8 registers in a house, each 8 in.
by 10 in. How large should the cold-air box be ?
19. The sectional area of a cold-air box is 600 sq. in. ; each
register measures 8 in. by 10 in. How many registers are there
in the house ?
20. The cross section of a cold-air box measures 20 in. by 20
in. There are 6 equal registers in the house. What is the area
of each ?
21. It is said that the minimum dimensions of an ideal din-
ing room are 11 ft. by 13£ ft. How many square feet are con-
tained in such a room ?
22. An ideal dining room of maximum size measures 17 ft.
by 22 ft. How many square feet are contained in such a room ?
23. A kitchen, according to one authority, should measure
10 ft. by 12 ft. If the range measures 2 ft. by 3 ft. 14 in., the
sink 1 ft. 6 in. by 3 ft. 15 in., two cupboards each 1 ft. 8 in. by
6 ft. 4 in., and the work table 4 ft. by 5 ft. 7 in., how many
square feet are allowed for " walking " space ?
24. It is said that stairs are well proportioned when 2 times
the height of the riser, added to the tread, equals 24 in. The
riser of such stairs is 7 in. What is the tread ?
25. Measure the stairs in your home. How near do they
come to the ideal measurements ?
26. How large a piece of material must I have to make a bag
10" x 14" when finished, if I allow 2" for a heading and £" on
three sides for hems ? Use the width of the material for the
depth of the bag.
CHAPTER VI
Polynomials. Multiplication of Polynomials
103. Polynomial by Monomial. Review multiplication of
monomials, §§ 31-34. Give sign rule for multiplication.
Give exponent rule for multiplication. Give sign rule for
addition. Illustrate each.
104. In § 34, we found the product of a monomial by a
monomial. We shall now extend multiplication to cover any
number of terms.
A polynomial is simply a sum of monomial terms.
Hence, to multiply a polynomial by a number is to multiply
each of its terms by that number, and find the sum of these
products.
Ex. Multiply 5 a 2 + 3 ax - a? by 2 a.
6a 2 -2a = 10a 8 , Sax • 2a = 6a% — x* . 2a = — 2a3 3 .
Then, (5 a 2 + 3 ax - x*) . (2 a) = 10 a 8 + 6 a 2 x — 2 as 2 .
The work should be written in the following form :
5 a 2 + 3 ax-* 2
2a
10 a 3 + 6 a 2 x - 2 ax 2
Begin multiplication at the left.
EXERCISE 26
Find the following products. (Perform the numerical multi-
plication mentally, writing results oivVj.")
76
MULTIPLICATION OF POLYNOMIALS 77
1. 7a 2 + 3a6 + 26 2 3. -4:a? + 2xz-5z 2
5a —6a;
2. 5m 2 -3cm + y 4. - 14 ad 2 + 15 aH + 17 a*
2chj -13a
5. 18s 2 -17a#-24y s
12s 4
6. In example 5, substitute a? = l, y = 2 in your multipli-
cand, multiplier, and product. Is the result what you might
expect ?
Any example in multiplication may be checked by substitut-
ing some numerical value for each letter as suggested in exam-
ple 6.
7. Check each of the first five examples.
Multiply :
a 32 ah — 14 X2? by 16 ocz.
9. 115a^ 2 -112afy by -12.
10. 1024 a 2 + 612 db - 306 6 2 by 4 ab\
11. (aj + #) 2 + 6(a; + y) + 9 by (« + y>
12. (^ + y) 8 + 3(a; + y)-4 byl6(a + y) 2 .
13. -24(a?-2/) 8 + 14(a + &) 2 -21 by 16.
14. 3(z + y) 2 + 12(s + 2,) + 18 by4(* + y).
15. Check example 14.
105. Polynomial by Polynomial. To multiply a polynomial
by a polynomial is to multiply the multiplicand by each term
of the multiplier; and add the partial products.
Ex. 1. Find the product of 2 a + 3 6 by 3 a — 5 6.
(2 a + 3 6)3 a = 6 a* \ Q ab, VV&f^
(2 a + 3 6) ( - 5 6) = - 10 ab - Ifc b* . -
78 MATHEMATICS
Adding these partial products
6 a 2 + 9 ab
- 10 ab- 16b 2
we have 6 a 2 - ab - 15 b 2
The work should be written as follows :
2a + Sb
3a-66
6a 2 + 9ab
- 10 ab- 15 b 2
6 a 2 - ab-16 b 2
The procedure is the same for any number of terms. The
pupil will find the work more simple if both multiplicand and
multiplier are arranged according to either the descending or
ascending powers of some letter.
Ex.2. Multiply 5 a; -60* + or* -4 by -3a? + a; 2 -l.
Rearranging according to the descending powers of x,
x 8 -6x 2 + 5x-4
a? _ 3 x - l
x*-6x* + 5 a 8 - 4s 2
- 3 x* + 18 x 8 - 15 x 2 + 12 x
-x 8 + 6 a 2 — 5x + 4
x 6 -9x* + 22'x 8 -13x 2 + 7x + 4
EXERCISE 27
Find the following products. Check each result :
1. (5a + 3&)(2a+ 46). 6. (5^-4aj 2 -f 3aj-2)(2a;-7).
2. (6aj + 22/)(5z-3y). 7 - O 2 + » + 2) (a? - » 4- 2).
3. (7c + 2d)(7c-3d). a (a^ + a^ + ^X^-a^ + .y 2 ).
4. (7c + 2d)(7c-2d). 9. ( a 2 + 2 a& + ft 2 ) (a + 6).
5. (a? + «* + 1) (s - 1). 10. (a 2 -2ab + b 2 )(a-b).
Are the products in 9 and 10 alike ?
22. Multiply a 2 + 2a + 4 by a — 2.
2* Multiply 5x*- 30 a + 45 by 5 x -15.
MULTIPLICATION OF POLYNOMIALS 79
13. (a + 6)(a + &). 15. (x-2y)\
14. (a + 26)(a + 26). 16. (x-y)(x-y).
Note the form of these results in examples 13-16. They
will be useful later.
17. (a + b)(a-b). 19. (±x + y)(4x-y).
la (a + 26)(a-26). 20. (±x + 3y)(±x-3y).
21. (16c + 15d)(16c-15d).
Note the form of the results in examples 17-21.
Write the following products by inspection :
22. (c + 2d)(c-2tf). 25. (3x + 2y)(3x + 2y).
23. (5x + y)(px-y). 26. (2a + 3c)(2a + 3c).
24. (5x + y)(5x + y). 27. (5m + 4ft)(om + 4A;).
Solve examples 28-32 mentally : *
28. The side of a square is a -f b. Find the area.
. 29. The side of a square is 2 a — b. Find the area.
30. The side of a square is 4 x + 3 y. Find the area.
31. The side of a square is 6 c — 4 d. Find the area.
32. The side of a square is 5 x — 2 a. Find the area.
33. In examples 28-32 find the dimensions and the area if
a = 2, 6 = 1, c = — 1, d = 3, x = 4, y = — 1.
Find the areas of the following A where b = base and a =
altitude :
b
a
34.
6x + 3y,
5x + 7y.
35.
a + b,
a — b.
36.
x + S,
x-2.
37.
x + 9y,
x+2y.
aa
x+15a,
x — $a.
80 MATHEMATICS .
39. Compute the areas in examples 34-38 when a = 3, 6 = 1,
x = 7,y = -6.
40. 16ir 4 4-(8ar J -3a; + 14)(5a;-2^-h24)-336 = ?
41. [(x* + 3 x -4) + (x* --3 x + lfti? x*--6x + l) = ?
42. (2a? + 3)(2a? + 3) 2 =?
43. The edge of a cube is 4 a -f 6; find its volume.
44. The edge of a cube is 2 a? — 5 y ; find its volume.
45. The edge of a cube is 5 x — 14 b ; find its volume.
46. Find the volumes in examples 43-45 if the letters have
the same values as in example 33.
Find the areas of the following trapezoids: (B represents
the lower base, b the upper base, and a the altitude).
B b a
47. x + y y < x-y, 2.
4a 3x + 2y, 2x + 3y, 4aj + 4#.
49. 5x + 2y, 3x — y, $x + y.
50. 3x + 5y, Ix — 3 y, hy + x.
51. 7c+3rf, 2c+d, c-d.
52. 4 c - 8 d, c+5d, c-3d.
53. x + 9y, x — 6y, x — y.
54. x -f 18, x — 15, 4y — x.
55. c + 8, c-5, c-2.
56. Find the upper base, lower base, altitude, and area in
examples 49-55, if a; = 3, y = 2, c = 4, d = — 5. How do you
account for your negative areas ?
57. If 10' be subtracted from the length of a rectangle, and
the same amount is added to the breadth, the area will be
increased by 100 square feet, but if 10' be subtracted from the
breadth and 10' added to the length, fne &Tfc& V\\\.\& &\\si\m%\\&&
MULTIPLICATION OF POLYNOMIALS 81
by 300 square feet. Make a diagram of each rectangle, using
a scale of T y = 1 foot.
Division of Polynomials
106. Polynomial by Monomial. Review division of mono-
mials, §§ 32-35. Give sign rule for division. Give exponent
rule for division. Give rule for subtraction of monomials.
107. Since a polynomial is a sum of monomial terms, to
divide a polynomial by a monomial, divide each term of the
polynomial by the monomial and add the quotients thus found.
Ex. 1. Divide 81 a 4 - 27 a 2 + 18 a by 9 a.
81 a 4 -s- 9a = 9a 8 , - 27 a 2 -^ 9a =- 3a, 18a -h 9a = 2.
Then (81 a 4 - 27 a 2 + 18 a) -- 9 a = 9 a 8 - 3 a + 2.
The work should be written as follows :
9 q )81 a 4 - 27 a 2 + 18 a
9a 8 - 3a + 2
Ex.2. Divide 21<K 4 -18» 8 + 5a 2 -9a; by -Sx.
-Sx ) 21 x 4 - 183? + 5a 2 -9a
-7x3+ 6 a 2 -fa +3
Ex. 3. Divide 4 a 2 (2 m+3)-9 a(2m+3) +2 m+3 by 2m+S.
( 2m + 3 )4 a 2 (2 m + 3)- 9 a(2 m + 3)4- 2 m + 3
4 a 2 • — 9 a +1
Note that in dividing one number by another, we are simply
removing from the dividend the factors found in the divisor.
Ex. 4. 45 + 3.
45 = 3 2 • 5.
3)3 2 . 5
3 6
After removing the 3 of the divisor 3 • 5 ore> Y&ft. lot \ha o^vscX.
82 MATHEMATICS
Ex.5. 45 afy-*-9 a 2 .
46 x*y = 3 2 • 5 • x • x • x • y.
9x 2 = 3 2 xx.
3 2 • x • a Q3 2 » 5 • a; • a; • x • y
5 • x • y = 6 xy
If one carries this principle in mind no rule for division by a
monomial is necessary.
EXERCISE 28
1. 72afy-s-24afy.
2. 4(a + &) 5 ■+- 4(a + &) 3 .
3. (128 a* 6 - 80^ + 32)-*- 16.
4. 128 a 6 - (80 a? 4 + 32)-*- 16.
5. 128 a 6 - 80 a? 4 + 32 -«- 16.
6. (128 a 6 - 80 a? 4 ) + 32 --16.
7. [7a? 4 (2a+6) 3 -12ic 2 (2a+6) 2 +15a?(2a+6)]^-(2a+6).
8. Divide 14 a(x — y) + 7 a 2 (a? — y) 2 — 49 a\x — y) 3 by
-7a(a?-y).
9. Divide (a + 6)a 2 + (a + 6)2 a& + 6 2 (a + 6) by a + 6.
10. Divide 144 x A yh ~ 729 afy 8 * 4 + 162 xtfz? by - 3 afy 2 z.
108. Polynomial by polynomial.
The product ofa? + a) + 2by2a; + 3is found by
a 2 (2a; + 3) + a;(2a; + 3) + 2(2a! + 3) (1)
= 2a? + 3a 2 + 2a 2 + 3a; + 4a> + 6 (2)
= 2a* + 5s 2 + 7a? + 6. (3)
Ex. 1. Kequired to divide 2 X s + 5 x 2 + 7 x + 6 by 2ar + 3.
This means that 23 8 + 6x 2 + 7x + 6 is the product of two factors or
sets of factors, one of which is 2 x + 3. The problem is to find the other
factor.
If2o^ + 5a5 2 + 7a5 + 6is written in form (1), division can be performed
as in example 3, § 107, but if the multiplication has been completed and
the partial products added as in form (3), the factor required is not so
readily seen.
MULTIPLICATION OF POLYNOMIALS 83
It is evident from (1) that (3) is made up of partial products. If
2 x -f 3 is the divisor, x 2 + x + 2 is the quotient.
An examination of (1) shows that the first term of each partial product
is the product of the first term of the divisor by the corresponding term of
the quotient. Therefore we may form this rule :
1. Arrange both dividend and divisor according to the as-
cending or descending powers of the same letter.
2. Divide the first term of the dividend by the first term of
the divisor.
3. Multiply each term of the divisor by the quotient found
in 2.
4. Subtract the product found in 3 from the dividend,
arranging the difference found in the same order as the
dividend.
5. Divide the first term of the difference by the first term
of the divisor. This gives a second term of the quotient.
6. Proceed in this manner, considering each difference as
a new dividend until the first term of the difference is of
lower degree than the first term of the divisor.
7. If there is a remainder, make it the numerator of a frac-
tion whose denominator is the divisor, and annex with proper
sign to the quotient.
Dividend Divisor
2x* + 5x 2 + 7x + 6 2x + 3
1st partial product, x 2 (2 x + 3) = 2 X s -f 3 x 2 x 2 + x + 2
2 x 2 + 7 x + 6 Quotient
2d partial product, x(2 x + 3) = 2 x 2 + 3 x
3d partial product, 2(2 x + 3) = 4x + 6
Ex.2. Divide « 8 + 3a;-2 by a?-4.
a 8 + 3 x - 2
aj8-4a; 2
a — 4
74
z 2 + 4x+19 +
4x* + 3x-2 *~ 4
4 x 2 - 16 x
19x-2
10 x - 76
x — 4
84 MATHEMATICS
EXERCI8B 29
Verify each result :
1. Divide x*(2 x + 3) + 8 x(2 x + 3) + 15(2 x + 3) by 2 a?+3.
2. Divide a\a + ft) + 2 aft(a + ft) + (a + ft) 2 by a + 6.
3. The area of a rectangle is a? + 8 a? + 15. The length is
x + 5. What is the breadth ? What is the breadth if x = 2 ?
4. Divide y*-2y-15 by y + 3.
5. Divide y 2 + 2y-15 by y + 3.
6. Divide y* + 2y- 15 by y-3.
7. Divide y 2 - 8 # + 15 by y- 3.
8. (5« 8 -20a 2 + 15aj-30)-5--5.
9. 5x*-(20x* + 15x-30)+-5.
10. 5« 8 -20i» 2 + (15aj-30)-5--5.
11. 5x*-20x* + 15x-30 + 5 + 3(2x-8)-}-2.
12. 4(a + ft) 2 -J- 2(a + ft) + 4 (a + ft) 2 -*■ - 2(a + &).
13. The radius of a circle is x + 3. Find the area.
14. The area of a trapezoid is 2 sc 2 + 12 x + 18 ; the sum of
the bases is 4 # + 12. Find the altitude. If the upper base
is x + 7, what is the lower base? If x = 3, what are the
dimensions? Can you draw the trapezoid if it is isosceles?
15. Divide x* + 3s 2 + 3a + l by x* + 2x + l.
16. Divide x s -{-3x 2 + 3x + l bya + 1. Divide the quotient
by x + 1. Compare results in examples 15 and 16.
17. Divide X s + 27 by x + 3. 18. Divide a 3 + 1 by a + 1.
19. Divide 8 a 3 - 27 ft 3 by 2a-3ft.
20. Divide x 5 - 32 by x - 2.
21. Divide a? 4 - 81 by x - 3.
22. Divide x 4 + 81 \>y x -V S.
MULTIPLICATION OF POLYNOMIALS 85
23. Divide x 2 - x - 72 by x - 9.
24. The area of a trapezoid is 15 x 2 — 34 x -f- 16 ; the altitude
is 5 x — 8 ; one base is 2 x -f 3. Find the other base.
25. The area of an isosceles triangle is 12x* + 32 x — 35;
the base is 6 a? — 5. Find the altitude, then construct the
triangle when x = 2. Construct the triangle when x = — 2.
26. The area of a rectangle is 4 x 2 + 12 a; + 9 ; the base is
2 a; + 3. Find the altitude. Construct the rectangle. Com-
pare the base and altitude of your drawing. What kind of a
rectangle is it?
27. Divide the sum of x 3 + 7 x 2 + 35 x + 19 and
x 3 -f 8 x 2 —13 x — 34 by the difference between 5 a? 2 +2 a?— 7 and
28. Perform the following operation: ( 3a? + 7)(a^- a?- 12) >
# — 4
29. ic 2 (2« 8 + 9aJ 2 -71a;-120)-h(ar 2 + 3aj-40>
30. a>«(2 a? + 9 a? - 71 a;-120)-s-ar> + 3a>--40.
31. Divide a 2 + 2 ab + 6 2 by a + 6. Is a 2 + 2 a& + b 2 a square
or a rectangle ?- Define a square.
32. (100 x 4 - 229 x 2 + 9) -5- (5 x - 1).
33. Divide 100 z 4 - 229 a 2 + 9 by 4 x 2 - 9.
34. Divide 100 x 4 - 229 a 2 + 9 by 25a? - 1.
35. Divide 100 x 4 - 229 x 2 + 9 by 5 a? + 1.
36. Divide 100 a; 4 - 229 x 2 + 9 by 2 x + 3.
37. Divide 100 x 4 - 229 a 2 + 9 by 2 a? - 3.
3a What are the factors of 100 a 4 - 229 x l + 9 ? of 4 ar* - 9 ?
of 25^-1?
39. Divide a^ + Oa^-aj-SObya;-^
40. Divide a 8 + 6 a 2 - a; - 30 by a? + 3.
41. Divide a? + 6a? 2 - a? -30 by a + 5.
86 MATHEMATICS
42. What are the factors of <x? + 6a?— x — 30? What does
the product of these factors equal ?
43. The length, breadth, and thickness of a rectangular
solid are x + 5, x -f 3, and x — 2, respectively. Find its vol-
ume. What are its dimensions when x = 3? x = 2? x = l?
» = 0? x = -1?
44. Divide x* - 64 by x - 8.
45. Divide a? - 64 by x + 8.
46. What are the factors of x 2 — 64 ?
47. The volume of a circular cylinder is
(16 X s - 84 x 2 + 120 a? - 25)tt. The radius of the base is 2 a? -5.
Find the altitude. Find the dimensions when x = 3 ; 4 ; 5.
108. The various symbols of algebra enable us to express
many operations by using these symbols in place of words.
In other words, the symbols are the shorthand, the stenography,
of mathematics.
For example : If we wish to indicate the subtraction of
2x—l from 4 x + 7, we may write :
(a) from 4 x -f 7 take 2 x — 1 ; or we may write :
(6) (4* + 7)-(2s-l);
the two expressions being identical, and read in the same way.
In the next exercise be sure you translate the algebraic lan-
guage into the English before attempting to solve the problems.
REVIEW
1. (2s + 3)(2a>-7)+(3z-l)(2a>+5).
2. (2a + 3)(3s + 2)-(4<c + 15).
3. (4»-l)(» + 4)-(aj-4)(4* + l).
4. aj 8 -[3ar J +(3aj-l)]-[ic 3 +3» 2 -(3a;-l)].
A 4(*-5X*-2)-4.
MULTIPLICATION OF POLYNOMIALS 87
6. 7 -f 3 -r- 3 -4.
7. 2 +(8-s-4). 5 -8.
8. 6-90-s-(3.10) + 2-8.
9. (2x — 7)x-(3x + 4:)x.
10. Two right triangles are equal if two legs of the one are
equal respectively to two legs of the other. Prove.
11. In a rectangle the opposite sides are equal. Prove that
the diagonal divides a rectangle into two equal triangles.
12. The bisectors of the base angles of an isosceles triangle
are equal. Prove.
13. [(a + b) 5 + 6 b(a -f b) A - 12(a -f fc) 8 ] -5- (a + b) 2 .
14. (4a f -&*)-(2a + &)(2a-&).
15. (64c 2 ^-25d 2 )-(8c + 5d)(8c-5d).
16. (5x + 2y) 2 -(5x-2y)(5x-2y).
17. (5x + 2y)(5x-2y)-(25x 2 -4:y 2 ).
18. (7a>-33) 2 -(7a; + 32) 2 .
19. (ar + 5)(aj-2)-(aj-5)(a? + 2)-(»-5)(a;-2)
-(a + 5)(s + 2).
20. (a + &) 8 .
21. (c + 7) 8 .
22. (a 2 + a& + b*)(a 2 - a& .+ b 2 ).
23. (a 2 + a& + 6 2 )(a 2 + a& + b 2 ).
24. Find the area of a trapezoid whose upper base, lower
base, and altitude are x-\-7, 2a?-h8, 3a? + 2 respectively. If
x = 1 and the trapezoid is isosceles, construct the trapezoid.
25. Find the prime factors of (25) 2 , (24) 2 , (72) 2 .
26. What are the prime factors of (12) 6 ? (144) 5 ?
27. What are the prime factors of 1728 ? of (39) 3 ?
2a Factor (81) 4 , (8) 6 , (16) 10 .
88 MATHEMATICS
29. Factor 27 - 81 . 729.
30. Factor 72 . 64 . 36.
31. Factor 45 Tpy 2 . How many prime factors in this num-
ber?
32. Factor 81 (a + b)\
33. Factor 27 (2 x — y) 8 . If x = 3 and y = 1, what are the
prime factors of this product? How do they differ from the
prime factors of (15) 3 ?
Supplemental Applied Mathematics
1. One egg weighs 2 ounces. 57% of the egg is white,
32 % yoke, 11 % shell. Find the weights of the whites, yolks,
and shells of one dozen eggs.
2. One egg weighs 2 ounces. The shells of 2 eggs weigh
one ounce. What per cent of the whole egg is the edible
portion ?
3. The edible portion of 2 eggs measures f of a cup. It
takes 9 whites of eggs to measure 1 cup. How many yolks of
eggs does it take to make 1 cup ?
4. Beaten whites of 4 eggs measure 2£ cups. What is the
per cent of increase in quantity of beaten over unbeaten whites ?
5. Beaten yolks of 3 eggs measure £ cup. What is the per
cent of increase in quantity of beaten over unbeaten yolks ?
6. One egg, yolk and white beaten together, measures 4
tablespoonfuls. How much greater is the increase in quantity
when yolks and whites of 4 eggs are beaten separately ?
7. According to Hutchison, experiments as to the difference
in time of digesting eggs cooked in various ways show that 2
soft " boiled " eggs leave the stomach in If hours, and 2 hard
"boiled" eggs leave the stomach in 3 hours. If 2 eggs are
eaten on each of 4 days a week, how many more hours' work a
month will the stomach have in digesting hard than soft
"boiled " eggs ?
MULTIPLICATION OF POLYNOMIALS 89
8. An egg was boiled for 3 minutes. After artificially
digesting for 5 hours in a pepsin solution, it contained 8.3 %
undigested protein. An egg was cooked in water at 180° F. for
5 minutes. It was entirely digested after 5 hours in a pepsin
solution. The edible portion of eggs contains 13.4 % protein.
Find the weight of undigested protein from 1 dozen eggs if they
are soft " boiled " rather than soft " cooked."
9. Steel is 7.83 times as heavy as water. Find the weight
of one cubic inch of steel.
10. A gas range has four burners, each of which burns .65
cubic feet per minute and two oven burners, each burning 1.032
cubic feet per minute. Find the cost per hour of running the
stove when all burners are on full, gas at 65 ff per thousand.
11. A house is heated by a gas furnace containing four
burners and two pilot lights. Each burner consumes one cubic
foot in two minutes ; each pilot one cubic foot in eight minutes.
Find gas bill for February at 30^ per thousand cubic feet.
The pilots burned all the time. Two burners burned from
6 a.m. to 10 p.m. and four burners were running two hours in
the morning and two hours in the evening each day.
12. According to one authority, a family of five living on
$2000 to $4000 per year should spend 25% of that sum for
food; 20 % for rent ; 15 % for operating expenses, such as fuel,
wages, etc., 15% for clothing; and 25% for higher life, i.e.
books, travel, charity, saving, and insurance. What amount
should be used for each item if a family has an annual income
of $3000?
13. A family living on $1000 to $2000 per year requires
25% for food; 20% for rent; 15% for operating expenses;
20 % for clothing ; and 20 % for higher life. If a family lives
on $ 1500 per year, what amount should be spent for each item ?
14. From a family income of $ 800 to $ 1000 per year, 30 %
should be spent for food ; 20 % for rent •, 10 °fo fo* o^Rra&x&%
expenses; 10% for clothing-, and. 20 fy ira \£\^s* ^^
90 MATHEMATICS
How much can be spent for each item when the income is
$900?
15. A family living on $ 500 to $ 800 per year should spend
45% for food; 15% for rent; 10% for operating expenses;
10 % for clothing, and 10 % for higher life. If a family's
annual income is $ 650, how much should be spent for each
item ?
16. From an annual income of less than $ 500, 60 % should
be spent for food ; 15 % for rent ; 5 % for operating expenses ;
10 % for clothing ; and 10 % for higher life. If the income of
a family is $ 425 per year, how much can be spent for each item
per year and month ?
17. The grocery bill of a family living on $ 3000 per year
should be 25 % of that sum ; of a family living on $ 700 a
year, 45 % of that sum. Find the difference in the monthly
grocery bill of each ?
18. 20 % of a twelve-hundred-dollar income should be spent
for clothing, and 15 % of a nine-hundred-dollar income. Find
the difference in the amounts spent for these items by two
families having these incomes.
19. There are 49 lb. flour in a fourth barrel, or an ordinary
sack of flour. 1 lb. flour measures 4 cups. If a family uses 5
loaves bread per week, and it takes 3J- cups of flour to make
one loaf, how many months will a sack of flour last ?
20. Find the cost of 4 loaves of bread, requiring 1 hour for
baking and containing the following : 31 qt. flour, 1 yeast cake,
2 tb. lard, 4 t. salt, 4 t. sugar. Flour costs $ 2.00 per one-fourth
barrel; yeast 2 per cake; lard 15^ per pound (2 c. in a
pound), the sugar and salt $.0024; gas 70^ per 1000 ft., the
oven burner burning 39 cu. ft. per hour.
(Tb. = tablespoon, t. = teaspoon, c. = cup.)
21. I buy 28-inch material for handkerchiefs. How many
.yards would I have to buy to make 9 dozen handkerchiefs, cut
14 inches square ?
MULTIPLICATION OF POLYNOMIALS 91
22. What is the size of the finished handkerchief if a ^-inch
hem is made on four sides ?
23. If I paid 2 cents a yard for stitching, how much would the
work on 6 dozen handkerchiefs cost ?
24. How much lace would it take to sew around 9 dozen
handkerchiefs if you allowed 4 inches extra on each handker-
chief for fullness ?
25. If the material cost 60 cents a yard, and lace 15 cents a
yard, how much would the handkerchiefs in problem 24 cost,
including 5 spools thread at 5 cents a spool and 2 cents a yard
for stitching ?
CHAPTER VII
Graphs, the Algebra of Lines. Parallels and their Uses
109. In §§ 43-49 we studied and solved simultaneous equa-
tions. We will now study these equations from a geometric
standpoint.
In § 25 we determined that if toward the right were posi-
tive, toward the left should be negative. Suppose we
measure our positive and nega-
tive values from two lines inter-
secting at right angles. Suppose
further, that horizontal measure-
ments shall be admeasurements
-X and vertical measurements be
^-measurements, upward being
positive and downward being neg-
ative.
M
Y r
Ex. 1. Find the point where x = 2, y = 3.
Measuring OM = 2 and MP = 3, we have the point P satisfying the
required condition.
EXERCISE 30
In the same manner locate the following points :
1. x = 4, y = 2. 5. x = — 1, y = — 4.
2. x = 3, y = 5. 6. x = — 1, y = 2.
3. x = 2, y = — 3. 7. x = 5, y = — 1.
4. #=— 1,0 = 4. 8. x=^,'y= < i %
92
GRAPHS, THE ALGEBRA OF LINES
93
9. x = 2, y = 0. 10. x = 0, y= — 3.
11. x=-2, y = 0.
12. Are these points on the same circumference ?
s = 3,y = 4; a,- = 4,y = 3; <e= — 3,,y = 4;
<e = 4,y= — 3; x=— 4,y = 3; a = 3,y=— 4;
aj=— 3, y= — 4; a? = — 4, y = — 3.
(Take your center at 0.)
110. In the equation a? + y = 4, a; and y are so related that
their sum must always equal 4. x and y then may take any
values whose sum is 4. To find such pairs of values, solve
x + y = 4 for either sc or y. We will solve for y.
y = 4 — a?.
Give x a Bet of values, say 0, 1, 2, 3, etc., and find the corresponding
values of y. If x = 0, y = 4 ; if x = 1, y = 8, etc.
It is convenient to write these values in two columns headed x
and y. See below.
x
1
2
3
4
6
1
y
4(.4)
3(5)
2(C)
1 (2>)
0(2?)
1(F)
5 etc.
|T
^4
b
G
B
T-'
E
-A
M
«
r
t
Now plot these values on the same pair of axes, just as we
did in exercise 27.
In this way we obtain the points A, B, (7, 2), etc.
If these points are connected by a smooth curvfc., we<j ^\sA» ws. \&a
curve will correspond to a point satisfying an x &tA ** ^ oi x -V >& "=- ^~
94 MATHEMATICS
The values of x and y which locate the position of a point
are called coordinates of a point. The admeasurement, e.g., OM,
is called the abscissa, the ^-measurement, e.g., MB, is the ordi-
nate. The point is called the origin.
If each term of the equation is of the first degree, the
curve ABGDEF is a straight line and the equation is called
linear.
This curve is called the graph or geometric picture of the
equation.
111. The degree of a term is determined by the sum of the
exponents of the letters in it. In an equation the term of
highest degree determines the degree of the equation.
x*y + xy 2 — x 8 is an expression of the fourth degree in xy and
of the third degree in x.
x 3 — 3 x 2 = 8 is an equation of the third degree.
2 x + a 2 = b 3 — c is an equation of the first degree in x.
EXERCISE 31
Find the graphs of the following equations :
1. x + y = 6. 5. 2x — y = 5.
2. 2x + y = 8. 6. 3x — 2y = 5.
3. x — y = 4. 7. x^-4y = — 3.
4. 4:X — 3y = 12. 8. #+2# = -4.
112. If a pair of simultaneous equations (§ 43) are plotted
on the same axes, their graphs will usually intersect. In this
case the coordinates of the point of intersection are the same
as the values that are found for x and y when the equations
are solved.
This explains in a geometric way the name simultaneous
equations. The x and y must at the same time have values
which satisfy both equations.
GRAPHS, THE ALGEBRA OP LINES
95
Ex. Solve by grapl
is:
x — 2y =
-5,
(1)
x + y =
1.
(2)
From (1)
From (2)
X
y
X
y .
+4
i
1
+3
1
2
i
2
~i
-1
2
-1
2
*y
Y
,<$$
^»
iSTy*
^>r
*><'
"V
A
A
\
J>
Y'
In the above problem the lines
(1) and (2) intersect at x = — 1, y=2. If the equations are solved
by the principles of §§ 46-48, the same results are obtained.
113. Note that in these equations x and y vary. Any
change in one of these numbers causes a change in the other.
For this reason x and y are the variables in the equation. (Com-
pare § 44.) Letters and numbers whose values do not change
in an equation are called constants. The number which is inde-
pendent of the variables is called the absolute term.
Sometimes the graphs of equations will not intersect. The
equations are then said to be inconsistent.
Ex. Plot the equations x -f y = 4,
x + y = 2 .
The graphs of these equations do not intersect. This is be-
cause the lines are parallel.
114. Parallel Lines. Two straight lines are said to be parallel
when they have the same direction. It is evident that if they
are drawn through two different points they are everywhere
equally distant and can never meet.
115. Some of the graphs of equations do not intersect because
they coincide.
96 MATHEMATICS
Ex. Plot 2a; -f 3y = 3,
4 a? + 6y = 6.
These graphs coincide. Equations of this kind are said to be
equivalent. See example 2, exercise 18.
Equivalent equations and inconsistent equations have no
solution.
EXERCISE 32
Solve by means of graphs. • In each case solve algebraically,
also compare the result with the graphical solution. Note that
if the roots are fractional, the graphical solution is only ap-
proximately correct.
1. x + y = 7, 5 x_y = 2 7. 4a;-7y = 18,
2a;-3y=-6. 2 4 ' 6a; + 5y=-4.
2. 3a; + 4y = ll, 3z_y = 19 a 9aj-4y = l,
4ar-3.y=-2. 4 3 6* fia-f 2y = 3.
3. x + 2y=13, 6 . 5a; + 8y = 13, 9. 5a-4y = 2,
2x- y = l. 6a?-5y = l. 4a?+5y=-2.
4. x + 2y = $,
2x+ y = 4.
What angle is formed by the lines in examples 2, 3, 9 ?
10. 2x — 5y = 3, 12. 2x + y = 4:, 14, 4m-7v= — 3,
3s- 7^-y= 4jr. 2x + y = 6. u + 3 v = 4.
11. 8a; + 2y = l, 13. 5a;-3y = 4,
4 x -f y = j-. 5 a; — 3 y = 5.
What relations exist between the three lines :
15. 4a; + 3y = 2,
3 x — 4 y = 5,
3a?-4y = 8.
PARALLELS AND THEIR USES 97
16. 5t+ w = ll, 17. 2m + 3fc = 19,
3t-2u = 4. 3m -Ik = 3.
18. 6v — 8w = 5,
4 v -h 5 w = 7.
Parallels and their Uses
116. Parallel lines (§ 114) are of great importance in
geometry.
117. We assume that but one line can be drawn through a
given point parallel to a given line.
118. Prove that in the same plane two perpendiculars to the
same line are parallel.
Hint. Draw the line * + y = — 4. At points x = — 4, y = 0, and
x = 0, y = — 4, erect perpendiculars to this line. To prove that these
perpendiculars are parallel, suppose that they meet if produced, then
read Theorem IX. The method of proof you use here is called Indirect.
(§ 102.)
Compare your given equation, x — y = — 4, and x — y = 4,
with the three lines of your figure. Are similar relations
found in exercise 32 ?
Problem
119. To draw a straight line parallel to a given straight line.
1. Draw an indefinite line AB.
2. Choose a point P without line AB through which to
draw the parallel.
3. Draw AP and extend it to some point K.
4. At P construct an angle equal to Z A.
5. PZ, one side of Z ZPK is t\\e Yec^Ytfe^ ^sm&s^
98 MATHEMATICS
Theorem XIV
120. Two lines parallel to the same line are parallel to each
other.
Draw line a. Draw lines b and c parallel to a. We then have
Giver two lines b and c parallel to line a.
To prove line b || line a
Proof. 1. If b and c are not parallel they will, if produced,
meet at some point P.
2. The rest of the proof is left to the pupil. (See § 117.)
Theorem XV
121. A line perpendicular to one of two parallels is perpendicu-
lar to the other.
Draw line a || line b. Draw line c _L line b, intersecting line
a at K. We now have
Given lines a and b parallel, and line c JL line b.
To prove line c ± line a.
Proof. 1. Draw line m through K perpendicular to c.
2. Then, lines b and m are ± to c, and b and m are ||. (§ 118.)
3. Then, m and a coincide. (§ 117.)
4. Hence, cJLa.
122. The angle opposite the base of a triangle is the vertical
angle. (Any side may be the base.)
The vertex of the vertical angle is the vertex of the triangle.
The altitude of a triangle is the perpendicular from the vertex
to the base.
An exterior angle of a triangle is the angle formed by any
side of a triangle and the adjacent side produced.
EXERCISE 33
1. In an isosceles triangle draw the exterior angle at the
base : at the vertex.
I 2. Draw three altitudes of an isosceles tn&&^&.
PARALLELS AND THEIR USES 99
3. Draw three altitudes of an equilateral triangle.
4. Draw three altitudes of an obtuse angled triangle.
5. Draw three altitudes of a right triangle.
6. Draw the three bisectors of the angles of the triangles in
examples 2 to 5.
7. Do the three altitudes of these triangles ever meet in a
point ? Do they always meet in a point ?
8. Do the altitudes and the bisectors ever coincide ? If so,
when?
9. Do the three bisectors ever meet in a point ? Do they
always meet in a point? Later in the course you will be
called upon to prove your above conclusion.
123. If two lines AB and XFare cut by a third line MN 9
MN is said to be a transversal.
This transversal makes with the other two lines eight angles
which have special names,
names which refer to the
position of the angles
with respect to the lines.
For example, a, 6, o, s,
are between lines AB
and XT, and are called
interior angles. The re-
maining four are exterior
angles.
a and s being on oppo-
site sides of the trans-
versal are alternate-interior angles. Likewise o and b are
alternate-interior angles.
c and s, on the same side of the transversal are exterior-
interior angles. Locate the other exterior-interior angles.
d and g are alternate-exterior angles. Find the other alter-
nate-exterior angles.
100 MATHEMATICS
Theorem XVI
124. If two parallel lines are cut by a transversal, the alternate-
interior angles are equal.
Draw AB II XT, and MN intersecting AB and XTdX O and
K, respectively. Let Z KOA = c, and Z TKO = d. We
then have
Given lis AB and XFcut by MN forming alternate-interior
angles c and d.
To prove Z c = Z d.
Proof. 1. Through Z, the middle point of OK, draw a _L to
AB, meeting AB at F, and XT at J£. Let Z ^0 = ^ and
Z 7jTZ# = t.
2. JfrF± XF. (§ 121.)
3. In rt. A OFZ and KEZ,
9 = i. (§91.)
KZ= ZO. (Constr.)
4. Hence, AOFZ=AKEZ. (?)
5. Then, Zc = Zd. (§ 79.)
Tfcis is the fundamental proposition in parallel lines.
EXERCISE 34
1. If two parallels are cut by a transversal, the exterior-
interior angles are equal. (Use the equations derived from the
Theorems XVI and VII.) Note : Remember in proving any
geometric statement you must give authority (the why) for
each step you take.
2. If two parallels are cut by a transversal, the sum of the
interior angles on the same side of the transversal is equal to
two right angles.
3. If two parallels are cut by a transversal, the alternate-
exterior angles are equal.
PARALLELS AND THEIR USES 101
4. If two parallels are cut by a transversal, the sum of the
exterior angles on the same side of the transversal is equal to
two right angles.
Theorem XVII
125. If two lines are cut by a transversal, and the alternate-
interior angles are equal, the lines are parallel.
Draw a line AB. Draw a line MN intersecting AB at 0.
Through any point iTon MN, draw line XY, making an angle
YKO equal to KOA. Use same notation for angles as that
in Theorem XVI. We have
Given two lines AB and XFcut by MN making A c = Z d.
To prove AB II XY.
Proof. 1. Through K draw GH II AB.
2. Then, A OKH = Ac. (Theorem XVI.)
3. But Z d = Z c. (By Hyp.)
4. Then, A OKH=Z'd. (Ax. 8.)
5. Hence, lines Oil and XY coincide.
6. Therefore, X Y II AB.
This theorem is the converse of Theorem XVI. That is, the
hypothesis and conclusion of the two theorems are inter-
changed. Theorems XVI and XVII are very important.
EXERCISE 35
1. If two lines are cut by a transversal and the exterior-
interior angles are equal, the lines are parallel.
2. If two lines are cut by a transversal, and the sum of the
interior angles on the same side of the transversal is equal to
two right angles, the lines are parallel.
3. If two lines are cut by a transversal, and the alternate-
exterior angles are equal, the lines are parallel.
4. If two lines are cut by a transversal, and the sum of the
exterior angles on the same side of the transversal is 180°, the
lines are parallel.
102 MATHEMATICS
5. If two angles have their sides parallel, each to each, they
are either equal or supplementary. (Hint. Produce one side
of each angle, if necessary, until the lines intersect.)
*
6. If two angles have their sides perpendicular, each to
each, they are either equal or supplementary.
Draw Z ABC, also Z KOM whose sides are perpendicular,
each to each, to the sides of
S G A ABC. We have
S^ j Given A ABC and KOM
^ A j ,,'' with side AB 1_ KO, and
D'' |\ To prove Z.B equal to or
I \af supplementary to Z KOM.
K
1 Proof. 1. * Through draw
DE JL OM, D and E being on opposite sides of ; produce
KO to G, and draw OHJL GK Call £GOE,&; Z HOE, i\
Z MOH,c; AKOM, 6; Z DOK, a.
2. Then DE II BC (?)
3. And OH II BA. (?)
4. Theref ore Z i = Z B. (?)
5. i + c = 90°. (?)
6. 6 + c = 90°. (?)
7. Then i = b. (?)
8. And ZB = Z b. (?)
9. If OK were drawn in the opposite direction, namely OG,
Z MOG would be the supplement of Z B.
7. Two triangles have their sides mutually perpendicular.
Show that they are mutually equiangular.
8. Two triangles have their sides mutually parallel. Show
that they are mutually equiangular. Are the triangles equal ?
Remember that all additional lines drawn in &&£\xrom\&X\& tatted lines.
PARALLELS AND THEIR USES 103
Theorem XVIII
126. The sum of the angles of a triangle is equal to two right
angles.
Given A ABC.
To prove that Z A + Z B + Z ACB = 2 rt. A.
Proof. 1. Produce side BC to K. Draw C^f II 2L4, and on
same side of BC as BA.
Let Z ^CB = x, ZACM= y, Z.MCK= z.
2. * + y+* = 2rt.A (§ 76, c.)
3. ZJB = s (?)
4. ZA = y (?)
5. a; = x (?)
6. Add equations 3, 4, and 5.
7. Compare equations 2 and 6.
8. Hence?
EXERCISE 36
1. Prove Theorem XVIII by drawing a line through the
vertex, parallel to the base. (Do not draw any other lines.)
2. Prove two right triangles equal if a leg and an acute angle
of the one are equal, respectively, to the leg and acute angle of
the other.
3. Prove that the exterior angle (§ 122) of a triangle is
equal to the sum of the two opposite interior angles. (You
will need this theorem very often.)
4. Prove that the exterior angle of a triangle is greater
than either of the opposite interior angles.
5. How many right angles can a triangle have ?
6. How many obtuse angles can a triangle have?
7. One angle at the base of an isosceles triangle is 36°.
Find the other angles.
104 MATHEMATICS
8. The vertical angle of an isosceles triangle is 120°. The
base angles are bisected. Find the angles formed by the
bisectors.
9. The angle formed by the bisectors of the base angles of
an isosceles triangle is 100°. Find the angles of the triangle.
10. One angle of a right triangle is 45°. Compare the legs
of the triangle.
U. (a) Construct an equilateral triangle.
(b) Construct an angle of 30°.
12. Construct an angle of 45°.
13. An exterior angle at the base of an isosceles triangle
formed by producing the base is 108°. Find the angles of the
triangle.
14. Could the exterior angle in example 13 be 89° ? 90° ?
91°? Why?
15. An exterior angle formed at the vertex of an isosceles
triangle by producing one of the legs is 129°. Find the angles
of the triangle.
16. Find the exterior angle at the base of an equilateral
triangle.
17. Prove 6 of Exercise 35 by producing MO to meet BO,
and producing BA to meet OK produced.
127. Develop the following synopsis :
Two triangles are equal if (a)
Two right triangles are equal if (a)
(P)
(P)
(d)
Keep this synopsis always in youx mind.
PARALLELS AND THEIR USES 105
Theorem XIX
128. If two angles of a triangle are equal, the triangle is isosceles.
Draw a line CD. At C and D construct equal angles. Let
the sides of these angles meet at P. Call PC, d, and PD, c.
We then have
Given A CDP with ZC=ZD.
To prove c = d.
Proof. 1. Draw, a perpendicular from P to the base.
2. Prove the A formed are equal.
Theorem XX
129. If two sides of a triangle are unequal, the angles opposite
are unequal, and the greater angle lies opposite the greater side.
Draw A DEF making DE > DF. Call BE, f, and DF, e.
We now have
M E
Given A DEF with / > e.
To prove Z EFD > ZE.
Proof. 1. On DE take DM— e, and draw FM.
Call A DMF, x, and Z MFD, y.
(Note in this construction that D is the angle not involved
in the statement, and from D we measure the distance DM.)
2. x = y. (Theorem III.)
3. ZEFD>y. (Ax. 7.) i
4. Z x > Z E. (Ex. 36, 4.)
5. .-. y> Z E.
6. .\ZEFD> ZE. (?)
106 MATHEMATICS
EXERCISE 37
1. The perpendicular is the shortest line from a point to a
line. (Draw the perpendicular and any other line from the
given point to the given line.)
2. Two isosceles triangles are equal if the base and one
base angle of one are equal, respectively, to the base and one
base angle of the other.
3. The hypotenuse in a triangle is greater than either leg.
4. Find the sum of the angles of a quadrilateral. (Theorem
XVIII.)
5. If the lines are drawn from a point within a triangle to
the extremities of one side, the angle included by them is
greater than the angle included by the other two sides. (Use
the figure of Theorem VI, and apply Exercise 36, Example 4.)
Theorem XXI
130. Any point in the bisector of an angle is equidistant from
the sides of the angle.
Draw A DEF. Bisect A E.
From P, or any point in the bisector, draw PM and PK Jl
ED and EF, respectively, meeting ED at M and EF at K.
Call PM, d x and PK, d 2 . We now have
Given Z.DEF, and PE bisecting DEF, also, d v and d 2 Js
from any point, P, in the bisector, to sides ED and EF,
respectively.
To prove d\ = d 2 .
Proof. Show that A EPK= A EMP.
131. This bisector is sometimes called the locus of points equi<
distant from the sides of the angle. A locus may be defined as
a point or line which fulfills conditions \m^o^^L \^cm \t^ no
PARALLELS AND THEIR USES 107
other point or line meeting these conditions. Thus, in § 130
no point not in the bisector will satisfy the conditions of the
theorem. The center of a circle is the locus of all points in a
plane equidistant from the circumference.
EXERCISE 38
1. Show^that every point within an angle, and equally dis-
tant from the sides of the angle, lies in the bisector of the angle.
2. The bisectors of the base angles of an isosceles triangle
form with the base an isosceles triangle.
3. The point of intersection of the bisectors of the base
angles of a triangle lies in the bisector of the vertical angle.
4. Find the point in the base of a triangle which is equi-
distant from the other two sides of the triangle. Is this point
ever the middle of the base ?
5. Show that if lines are drawn from the middle point of
the base of an isosceles triangle respectively perpendicular to
the legs of the triangle, two equal triangles are formed.
6. If from any point in the base of an isosceles triangle,
parallels to the legs are drawn, two isosceles triangles are
formed. Are the triangles ever equal ?
7. Show that in a right triangle, if one angle is 30°, the
hypotenuse is twice the shorter side. (Bisect the vertical
angle of an equilateral triangle.)
8. Find a point which is equidistant from two intersecting
lines.
9. If three lines intersect, but do not pass through the
same point, find a point, if any such exists, which is equi-
distant from all three lines.
10. Bisect the exterior angles at the base of a triangle, and
show that these bisectors meet in a point of the bisector of the
angle at the vertex of the triangle.
10& MATHEMATICS
U. Bisect the exterior angle at the vertex of an isosceles
triangle. Show that this bisector is parallel to the base of the
triangle.
12. Through the vertex of an isosceles triangle draw a
parallel to the base. Show that this line bisects the exterior
angle formed by extending one of the equal sides through the
vertex.
13. Through the middle point of one leg of an isosceles tri-
angle draw a parallel to the other leg. Through the vertex
draw a parallel to the base. Show that two equal triangles
are formed. How do you find the middle point of one leg ?
14. Find a point equidistant from two parallel lines. Is
there more than one such point?
15. If two lines intersect, the bisectors of two adjacent angles
formed are mutually perpendicular.
Theorem XXII
132. If two triangles have two sides of one equal respectively
to two sides of the other, and the included angle of the first greater
than the included angle of the second, the third side of the first is
greater than the third side of the second.
Draw A ABO and KLM, having sides AC(b) and OB (a)
respectively equal to MK(l) and ML(k), and Z 0>ZM.
We now have
Given A ABC and KLM, with
b = l, a = k.
ZC>ZM.
To Prove AB (c) > KL(m).
Proof. 1. Apply A KLM to A ABC so that I will coincide
with b, L falling at L'. Draw CO, bisecting Z L'CB, meeting
AB at 0. Draw OL'
£ AO+OL'>AU. (?)
PARALLELS AND THEIR USES 109
3.
In A CL'O and COB,
■
CV = CB.
(?)
■
CO = CO,
-
Z VCO = Z OCB. '
(?)
4.
Hence,
A CL'O = A COB,
(?)
and
OV = OB,
(?)
5.
Then,
AO + OV = AO + OB.
(?)
6.
Hence,
AO+OB>AV,
(?)
or
c>m.
The converse of this theorem is also true. State the converse.
Four-sided Figures
133. A quadrilateral is a portion of a plane bounded by four
lines.
If the opposite sides are parallel, the figure is a parallelogram.
Draw the figure.
If two sides are parallel and the other two sides not parallel,
the figure is a trapezoid. Draw a trapezoid. In an isosceles
trapezoid the non-parallel sides are equal.
If no two sides are parallel, the figure is a trapezium.
A rhomboid is a parallelogram whose adjacent sides are not
equal and whose angles are oblique. Illustrate.
A rhombus is an equilateral parallelogram whose angles are
oblique.
A rectangle is a parallelogram whose angles are right angles.
A square is an equilateral rectangle.
The diagonal of a quadrilateral is a line drawn from one
vertex to the opposite vertex.
Begin lettering a four-sided figure at the lower left-hand
corner and read counter-clockwise.
Make a synopsis classifying quadrilaterals under three gen-
eral heads.
1 10 MATHEMATICS
Theorem XXIII
134. In a parallelogram the opposite sides are equal, and the
opposite angles are equal.
Draw a parallelogram ABCD. Draw BD. Prove the two
triangles formed are equal. Then use § 79.
This is the fundamental theorem in parallelograms.
EXERCISE 39
1. The diagonal of a parallelogram divides it into two equal
triangles.
2. The diagonals of a parallelogram bisect each other.
3. Parallel lines included between parallel lines are
equal.
4. From two opposite vertices of a parallelogram draw per-
pendiculars to the diagonal drawn between the other vertices.
Show that two pairs of equal triangles are formed.
5. Show that the lines connecting the middle points of the
opposite sides of a parallelogram bisect each other.
6. The diagonals of a rhombus bisect each other at right
angles.
7. In a certain parallelogram the diagonals bisect its
angles. One side of the parallelogram is 8'. Find the other
sides.
8. In an isosceles trapezoid draw perpendiculars from the
extremities of the shorter base to the longer base. Show that
two equal triangles are formed.
9. The diagonals of a rectangle are equal.
10. In an isosceles triangle ABC, the equal angles A and B
are bisected. These bisectors form an angle of 120°. The leg
of the triangle ABC is how many times its \ras&?
PARALLELS AND THEIR USES 111
Theorem XXIV
135. Two parallelograms are equal if two sides and the in-
cluded angle of one are respectively equal to two sides and the
included angle of the other.
Given fU ABCD and HEFG with AD, AB, and Z A respec-
tively equal to HG, HE, and Z H.
To prove O ABCD = a HEFG.
Proof. Draw diagonals DB and GE.
Prove A ADB = A HGE.
Then use exercise 39, 1.
Theorem XXV
136. If the opposite sides of a quadrilateral are equal, the
figure is a parallelogram.
Draw quadrilateral ABCD having AB = DC and AD = BC.
Draw BD and prove the triangles formed are equal.
Then use § 125.
EXERCISE 40
1. If two sides of a quadrilateral are equal and parallel,
the figure is a parallelogram.
2. If the diagonals of a quadrilateral bisect each other, the
figure is a parallelogram.
3. Two rectangles are equal when two adjacent sides of one
are respectively equal to two adjacent sides of the other.
4. The diagonals of a square bisect its angles.
5. If the diagonals of a rectangle bisect its angles, the rec-
tangle is equilateral.
112 MATHEMATICS
Theorem XXVI
137. If a aeries of parallels intercept equal parts on one trans-
versal, they intercept equal parts on every transversal.
Given lis AB, CD, and EF intercepting equal parts on line
GH.
To prove AB, CD, and EF intercept equal parts on every
transversal XY.
Proof. 1. Let HG cut AB, CD, and EF at M, N, K, respec-
tively, and XFcut these Us at M 1 , N', K*, respectively. Draw
M'O and N'R respectively II to MK.
2. MN=M'0 = NK=N'R. (?)
3. Now prove A 3fON' = A JSTRK'.
EXERCISE 41
1. A line which bisects one side of a triangle and is parallel
to another side bisects the third side. (In § 137, if M'K'
were drawn through M, we would have a triangle MKK 1 with
N the middle point on one side.)
2. The line parallel to a base of a trapezoid, and bisecting
one of the non-parallel sides, bisects the other non-parallel
side. (In § 137 KK'M'M is a trapezoid^
PARALLELS AND THEIR USES 113
Theorem XXVII
138. The line joining the middle points of two sides of a tri-
angle is parallel to the third side and equal to one half of it.
Draw A ABC. Through M and M\ middle points of AB
and AC, respectively, draw MM'. We then have
Given A ABC with MM' connecting the middle points of
the two sides.
To prove MM' II BC
Proof. 1. Through M draw a line parallel to BC.
2. This line will bisect AC and must therefore pass through
M 1 . Why ?
3. The parallel drawn will coincide with MM\ therefore
MM Ml BC.
4. Draw M'X II AB meeting BC at X.
5. Prove A AMM' = A M 'XC.
6. Then, MM' = XC = i BC.
Theorem XXVIII
139. The line joining the middle points of the non-parallel
sides of a trapezoid is parallel to the bases and equal to one half
of their sum.
Draw trapezoid ABCD, AB and DC being the parallel sides.
Draw EF joining E and F, the middle points of AD and BC>
respectively. We then have
Given trapezoid ABCD, with EF joining the middle points
of the non-parallel sides.
To prove EF II AB and DC, also EF= i(AB + DC).
Proof. 1. A line through E II AB will pass through F.
(Exercise 41, 2.)
2. Then EF II AB and DC.
3. Draw DB, intersecting EF at M.
4. In A ABD, EM is drawn through E the middle point of
AD and parallel to AB. (Use exercise 41, 1.)
5. Use A BCD in a similar maimer.
1 14 MATHEMATICS
Theorem XXIX
140. The bisectors of two of the angles of a triangle intersect
on the bisector of the third angle.
Draw A ABC, also BM bisecting Z B and CM bisecting
Z C. We now have
Given A ABC, and BM and CM bisecting A B and C, re-
spectively.
To prove that M lies in the bisector of Z A.
Proof. 1. Draw AM.
2. Since BM bisects Z B, M is equally distant from AB
and BC. (§ 130.)
3. Since CM bisects Z.C,Mis equally distant from BC and
CA. (§ 130.)
4. Therefore Mis equally distant from AB and AC.
5. Then M lies in the bisector of Z A, and AM is the
bisector. (Exercise 38, 1.)
141. From theorem XXIX we may assume that the three
bisectors of the angles of a triangle meet in a point, and the
point of intersection is equally distant from the three sides.
EXERCISE 42
1. In a parallelogram ABCD, M, the middle point of AB, is
joined to D, and M' , the middle point of DC, is joined to B.
Show that the diagonal AC is trisected.
2. Show that if the middle points of the sides of a quadri-
lateral are joined in order, the figure formed is a parallelogram.
3. If the middle points of the sides of a rectangle are
joined in order, the figure formed in a rhombus.
4. The figure formed by joining the middle points of the
sides of a square taken in order is a square.
5. Show that the two bisectors of the interior angles on the
same side of the transversal of two parallel lines form a right
angle.
6. What figure is formed by joining the middle points of
the sides of an isosceles trapezoid taken \w OT<tert
PARALLELS AND THEIR USES 115
7. Show that the opposite angles of an isosceles trapezoid
are supplementary.
Theorem XXX
142. The perpendiculars erected at the middle points of two
sides of a triangle meet in a point which lies in the perpendiadar
bisector of the third side.
Draw the perpendicular bisectors of two of the sides. Join
their point of intersection to the middle point of the third side.
Use § 94. Prove in a manner similar to that used in § 140.
143. We may conclude that the three perpendicular bisec-
tors of the three sides of a triangle meet in a point equidistant
from the three vertices.
Theorem XXXI
144. The three altitudes of a triangle meet in a common
point.
Draw A ABC, and its three altitudes x, y, z. We now have,
Given A ABC and the altitudes x, y, z.
To prove that x, y, and z have a common point.
Proof. 1. Through A draw MK II BC\ through B, draw
MH II AC; through C draw HKW BA.
2 x J_ MK. (§ 121.)
3. Since MBCA and ABCK are parallelograms, BC = MA
= AK.
4. Hence x is the perpendicular bisector of MK.
5. Similarly y is the perpendicular bisector of HK and z of
MH.
6. But the perpendicular bisectors of the sides of A MHK
meet in a point. (143.)
7. But 05, y, z are also the altitudes of A ABC
8. Therefore the altitudes of a triangle wis&fc ycl *» ^h&xs&tcl
point
116 MATHEMATICS
Theorem XXXII
145. Two medians * of a triangle meet in a point of t7ie third
median.
Draw A ABC, and medians CM and AK meeting at O. Also
draw BL through 0, meeting AC at L. We now have
Given A ABC with medians AK and CM meeting at O.
To prove that is in the median drawn to the third side.
Proof. 1. Draw CH parallel to AK and meeting BL pro-
duced at H. Draw AH.
2. In A HBC, OK II HC and bisects BC It therefore bisects
BH. (?)
3. Since is the middle point of HB, and M the middle
point of AB, MO II AH. (?)
4. Then AOCH is a parallelogram, and L is the middle
point of AC Why ?
5. Hence BL is a median and lies in the median.
146. In § 145, since HO = OB and LO = | #0, is f the
distance from B to Z. Similarly we may show that is % the
distance from C to ilf, and from A to X. Or the medians meet
in a point f the distance, along the median, from the vertex to
the opposite side.
Polygons
147. A polygon is a portion of a plane bounded by three or
more straight lines.
The bounding lines of a polygon are its sides.
Any two adjacent sides form an angle of the polygon.
A polygon is named with reference to its number of sides,
or angles. Thus, triangle, quadrilateral, pentagon, hexagon,
octagon, etc.
* A median is a line drawn from a vertex, oi a txtacL^le to the middle point
of the opposite side.
PARALLELS AND THEIR USES 117
If the sides are equal, the polygon is equilateral.
If the angles are equal, the polygon is equiangular.
A regular polygon is both equiangular and equilateral.
148. Two polygons are mutually equilateral if their corre-
sponding sides are equal.
Two polygons are mutually equiangular if their correspond-
ing angles are equal, that is, if their angles taken in the same
order are respectively equal.
Two polygons are equal if they are both mutually equilat-
eral and mutually equiangular. They are also equal if they
can be separated into the same number of triangles, equal each
to each, and similarly placed.
Theorem XXXIII
■
149. The sum of the angles of any polygon is equal to twice
as many right angles as the polygon has sides, less four right
angles.
Draw polygon ABODE, etc. We have
Given polygon ABCD, etc., having n sides.
To prove A i+Z5+ZC+ZJ9 + .-.etc., =2rirt. .4-4 rt. A.
Proof. 1. From any point P within the polygon draw a
line to each vertex.
2. It is evident that n triangles are formed, one for each
side of the polygon.
3. The sum of the angles of each triangle is two right angles,
and the sum of the angles of all the triangles together is 2 n
right angles.
4. The sum of the angles of all the triangles includes the
sum of the angles of the polygon, and the angles around P
which do not belong to the polygon.
5. Subtracting the sum of the angles at P from the sum of
the angles of the n triangles, we have 2 n rt. A — 4 rt. A for the
angles of the polygon. (§ 76, d.)
118 MATHEMATICS
Theorem XXXIV
150. The sum of the exterior angles of a polygon formed by
producing one side at each vertex of a polygon is equal to four
right angles.
Given a polygon with exterior angles b, d, f etc.
To prove that the sum of these angles is equal to 4 rt. A.
Proof. 1. Let a be the interior angle of the polygon adja-
cent to b.
2. Thena-f 6 = 2 rt. A.
3. This same sum exists at each of the n vertices, therefore
the sum of the interior and the exterior angles of the polygon
is 2 n rt. A.
4. But the sum of the interior is 2 n rt. A — 4 rt. A. (?)
5. Subtracting the sum of the interior angles from the sum
of the interior and exterior angles, we have
2 n rt. A-(2nrt.A-4:Tt.A),
or 4 rt. A.
EXERCISE 43
1. Prove Theorem XXXIII by triangles formed by drawing
all the diagonals from any one vertex of the polygon.
2. One angle of a regular polygon is 1^ right angles. How
many sides has the polygon ? How many degrees is the sum
of its angles ?
3. One angle of a regular polygon is 120°. What kind of a
polygon is it?
4. An exterior, angle of a regular polygon is 120°. How
many sides has the polygon ?
5. Each exterior angle of a polygon is 30°. Find the sum
of the interior angles.
6. One angle of a regular polygon is £ rt. Z. Find the
number of sides. A ny trouble Yieie ? "N^\^ ^
PARALLELS AND THEIR USES 119
7. From one vertex of a regular polygon 5 diagonals can be
drawn. Find the sum of the angles of the polygon.
8. What three lines belonging to any triangle meet in a
point ? Is there more than one answer to this question ?
9. One exterior angle of a regular polygon is 180°. How
many sides has the polygon? What is the greatest exterior
angle a regular polygon can have ? What is the least exterior
angle possible ?
151. Develop the following synopses :
Two lines are equal, if
Two angles are equal, if
Two lines are parallel, if
A quadrilateral is a parallelogram, if
Supplemental Applied Mathematics
1. A hexagonal water tank is 3' — 6" on a side. Find the
area of the cover.
2. In nuts and heads of bolts, the distance across the flats
(the distance between the parallel sides) is f the diameter of
the bolt plus \". The thickness of the head is £ the distance
across the flats. Find the distance across the flats and the
thickness of the head on a one-inch bolt having a square
head.
3. A hexagonal head is $" on a side. Find the distance
across the flats and the thickness of the head.
4. A square head ^" thick is to be milled on a cylindrical
blank. Find the diameter of the blank.
5. In example 3, find the approximate diameter of the
bolt.
6. A half-inch bolt has a square head. Find the distance
across the flats, the thickness of the head and the diagonal of
the head.
120 MATHEMATICS
7. A one-inch steel bolt is 4" long under the head. The
head is square. Find weight of these bolts per hundred.
8. A square head measures \" across the flats. The bolt
is 5" long under the head. Find weight per hundred.
9. The thickness of a hexagonal steel head is f ". Find
weight of heads per hundred. What size bolt would you use
with such head ?
10. A hexagonal head is £ " on a side. Find the diameter of
the bolt.
11. A hole is 0.185". Could you use a -fa" bolt in this hole ?
Would a \" bolt be too large ?
12. A hole is 1.284". The longest diagonal of the hexag-
onal head used is 2.74". Find size of bolt used, weight of nuts
per hundred, number of nuts per hundred pounds. (Bolts
only come in 4ths, 8ths, 16ths, 32ds, 64ths).
13. Whites of eggs coagulate at 56.6° C. Express this tem-
perature in Fahrenheit scale. Yolks of eggs coagulate at
122° F. Express in Centigrade.
14. When eggs are made into omelets, 1 tablespoonful of
milk and £ teaspoonf ul of butter are added to each egg. 7 eggs
will make 5 portions for serving. Find difference in cost of 6
eggs and enough omelet to serve six persons, when eggs cost
30^ per dozen, milk 6 ft per quart, butter 37^ per pound. One
cup of milk measures 16 tablespoonfuls, and one half pound of
butter 16 teaspoonf uls.
15. 87 % of milk is water. 1 cup of milk weighs 8 J ounces.
Find weight of water contained in one quart of milk.
16. 3.3 <f of milk is protein. £ of the protein is casein and J-
albumen. What is the per cent of casein and albumen in milk ?
17. The edible portion of cooked eggs contains 13.2 % pro-
tein. How much milk (liquid measure) does it take to contain
as much protein as is in one dozen egga ?
PARALLELS AND THEIR USES 121
18. A can of condensed milk costing 5^ contains £ of a cup.
Which is cheaper — fresh or condensed milk — if fresh milk
costs 6 ^ per quart ? (Dilute condensed milk one third.)
19. American cheese contains 28.8% protein. How many-
eggs contain as much protein as 1 pound of cheese? How
much milk (liquid measure) in one pound of cheese? How
much cheese contains as much protein as 2 eggs ?
20. Find difference in cost of 1 dozen eggs and as much
cheese as would contain the same amount of protein, cheese
costing 20^ per pound and eggs 30 ft per dozen.
21. One quart of sour milk makes 1 cup of cottage cheese
which weighs 6 ounces. Find difference in cost of 1 pound
cottage and 1 pound American cheese. (Prices same as above.)
22. Cottage cheese contains 20.9% protein. How much
cottage cheese contains as much protein as 1 pound American
cheese ? How much sour milk will be required to make this
quantity of cottage cheese ?
23. \ pound macaroni or 1 cup rice can be used with cheese.
1 cup rice weighs 7£ ounces, and costs 10^ per pound. Maca-
roni costs 15 ^ per pound package. Find difference in cost.
24. Boiled rice contains 24.4 % carbohydrates ; cooked maca-
roni contains 15.8 % carbohydrates. How many quarts cooked
macaroni will contain as much carbohydrates as 5 cups cooked
rice?
25. Uncooked macaroni contains 74.1 % carbohydrates;
cooked macaroni contains 15.8 % carbohydrates. Find weight
of loss in carbohydrates from cooking 1 pound macaroni.
26. I wish to make a dusting cap 18" in diameter. How
much lace is needed to put around the edge, allowing one half
extra for fullness ?
27. Two inches in from the edge of the cap in Example 26
I sew beading. How much beading mus&Wix^
CHAPTEK VIII
Products and Factors
152. In exercises 5 and 14 we found the factors of monomial
products. We shall now extend factoring to include the prod-
ucts found in exercises 27 and 29.
The Difference of Two Squares
153. Type I. Multiply a + bby a— b. (That is, multiply
the sum of two numbers by the difference of the same two
numbers.)
By actual multiplication : /
a + b
a — b
a 2 + ab
— ab —
6 2
a 2
62
That is, (a + b)(a - b) = a 2 - b 2 .
Or, stated in words :
The product of the sum and difference of two numbers is equal
to the difference of their squares.
Then to multiply the sum of two numbers, as 3 x + 5, by the
difference of the same two numbers, 3 x — 5, one needs only to
square the first number, 3x, and subtract the square of the
second number, 5, from it, giving for the product, 9 x 2 — 25.
EXERCISE 44
Find by inspection :
1. (x + y)(x — y). 3. (m -f- x)(m — x).
2. (x + 3)(x-3). *. (x-V^x-^Y
122
PRODUCTS AND FACTORS 123
5. (2a? + l)(2ar-l). a (14 y + 15)(14y - 15).
6. (2a? + 3)(2a>-3). 9. (17 a + 19 b) (17 a - 196):
7. (5s + 7)(5a;-7). 10. (16c + 25d)(16c - 25a*).
11. 0^ — 4 is the product of what two numbers ? Are these
numbers binomials, trinomials, or monomials ?
12. What are the factors of x 2 — 25.
13. Restate the rule in Type .1, making it applicable for fac-
toring such examples as example 12. Keep in mind that
factoring is a process of division ; division by inspection. The
dividend is given. You must find the divisor and the quotient.
Factor the following :
14. a 2 -25. 20. 16^cV-25arti 2 a 4 .
15. a 2 -81. 21. a? 4 -225.
16. a 2 -36. 22. 289 -a? 4 .
17. 25 a 2 -36. 23. 361s 2 -529/.
18. 16y 2 -25. 24. 441 y 4 - 729 a 4 .
19. 16 j/ 3 -25c 2 . 25. 64 a 2 - 196 b\
26. a 4 - 256 b* = (a 2 + 16 ft 2 ) (a 2 - 16 b 2 )
= (a 2 + 16& 2 )(a + 4&)(a - 46).
27. 16wi 4 -81c 4 . 31. 3 8 -2 8 .
28. (49) 2 -(25) 2 . 32. ( a +3) 2 -(a-3) 2 .
29. (561) 2 - (559) 2 . 33. (x + 7) 2 - (x - 7) 2 .
30. (625) 2 - (576) 2 . 34. (a + b) 2 - (a - b) 2 .
35. (2, + 21) 2 -0/-21) 2 .
36. (17x + 16y)(17x-16y) = ?
37. (21a+236)(21a-236) = ?
38. (24m + 19c)(24m-19c) = ?
39. (25a? < +18c)(25x-l%c)= c >
124 MATHEMATICS
40. (15c + 26d)(15c- 26tf) = ?
41. ^ a 2 -2£b*.
* 42. Can you factor a 2 + 25? Why ?
The square of the sum of two numbers.
The square of the difference of two numbers.
154. Type II. Multiply a + ft by a -+- ft. That is, multiply
the sum of two numbers by the sum of the same two numbers.
a + b
a + b
a 2 4- ab
+ ab + b*
a 2 + 2 ab + b*
That is, (a + ft) 2 = a 2 + 2 ab + ft 2 .
Or, stated in words : (a is the first number and b is the second number.)
The square of the sum of two numbers is equal to the square of
the first number plus twice the product of the first and second, plus
the square of the second.
Similarly, (a - ft) 2 = a 2 - 2 ab + ft 2 .
Or, The square of the difference between two numbers is equal to
the square of the first, minus twice the product of the first and
second, plus the square of the second.
EXERCISE 45
Find by inspection :
1. (a + z)(x + z). 7. (5a + 3) 2 . 13. (9m + 7a?) 2 .
2. (a> + 3)(a> + 3). 8. (5a-3) 2 . 14. (18 a -15 b) 2 .
3. (a -fa?) 2 . 9. (5a+4 6) 2 . 15. (5 m — a) 2 .
4. (a-xf. 10. (6y + 5c) 2 , 16. (6a -5b) 2 .
5. (a + 2x) 2 . 11. (7z-4z) 2 . 17. (5b-6a) 2 .
e. (a-2xy. 12. (8c- 7 d)*.
PRODUCTS AND FACTORS 125
18. From what do you get this product : x 2 + 6 x + 9 ?
19. From what factors do you get a 2 — 8 a + 16 ?
20. What are the factors of x 2 — 10 x + 25 ?
21. How can you distinguish a trinomial square ?
Find the factors of the following :
22. a 2 -6a + 9. 26. 9 x 2 + 30 x + 25.
23. a 2 — 4a + 4. 27. ra 2 + 12m + 36.
24. 4a 2 + 4a + l. 28. 4c 2 + 12c + 9.
25. 9» 2 -6a; + l. 29. 49 d 2 - 14 cd + c 2 .
30. ra 2 ^ 2 + 4 rafcc + 4 c 2 .
31. a 4 - 18 a 2 + 81 =; (a 2 - 9)(a 2 - 9) = (a + 3)(a - 3)(a + 3)
( a _ 3) by Type I.
32. x'-Sa^ + ie. 33. 16 c 4 - 72 cW + 81 d\
34. The area of a square is 25 a 2 + 40 ab 4- 16 6 2 ; find one
side of the square. Find the dimensions when a = 5, 6 = — 3.
35. One side of a square is 2x + 5y. What is the area ?
Find the area when x = 6, y = 1.
36. Is a 2 + 8 a + 25 a square ? Why ?
37. Is a 2 + 10 a + 24 a square ? Why ?
38. Is a 2 + 10 a + 25 a square ? Why ?
39. Draw a square whose side is a -\- b inches. Draw per-
pendiculars at the ends of a and b and show that this square
is made up of the square on line a, plus two rectangles each
of whose areas is ab, and a square on line b.
40. Factor a 4 — 16. Are your factors factorable ?
41. Factor x 4 — 81. What type are you using?
42. (81) 2 = (80 + l) 2
= 6400 + 160 + 1
= 6561
126 MATHEMATICS
Write the following squares by inspection :
43. (41) a 45. (79)* 47. (145) 2 49. (162) 2
44. (39) 2 46. (132)* 48. (153) 2 50. (169) 2
Trinomials of the form x 2 + kx + c. Binomial Factors having
One Term Common.
155.
Type
III.
Multiply x + 5 by
x +5
x + 3
« 2 + 5x
+ 3x4-15
x + 3.
•
x 2 + 8x4-15
Note that in this trinomial the first. term is the square of the
common term (x). The coefficient of x is the sum of the unlike
terms, 5 and 3. The third term is the product of the unlike
terms.
Ex. Find the product of 4 x + 3 and 4 x — 9.
Bytherule: (4x + 3)(4x- 9) = (4x) 2 4-(3 - 9)(4x) - 27-
= 16x 2 -24x-27.
EXERCISE 46
Find by inspection :
1. (a + 3)(a + 4). n. (p _ 14) (b + 16).
2. ( c + 5)(c + 7). 12. (c + 115)(c - 12).
3. (2a + 5)(2s + 3). 13. (d + 15)(d 4- 15).
4. (a + 4) (a -3). 14. (2a;-14)(2aj + 9).
5. ( a -4)(a + 3). 15. (2<e + 14)(2<e + 9).
6. (5a-4)(5a-4). 16. (2s + 14)(2a;-9).
7. (m + 18)(ra-15). 17. (y + 5)(# - 40).
8. (fc-18)(fc + 16). 18. (z + 27)(*-5).
9. (3fc + ll)(3fc-4). 19. (fc4-28)(fc-7).
la (c + 21)(c + 22). 20. (,a-lT^a-\5y
PRODUCTS AND FACTORS 127
21. From what factors do you get x 2 -f 7 x + 12 ?
22. Factor x 2 — x — 12.
To get the second terms of these binomials : factor the third
term into two factors whose sum is the coefficient of the unknown
in the second term.
Factor the following:
23. x 2 -2x-$. 33. y 2 -9y-36.
24. ^ + 6 x + 8. 34. a 2 -16 a + 64.
25. ar* — 6a + 8. 35. 4a 2 + 16a + 15.
26. a^-|>2a;-8.
(2 a is the term common to each
binomial.)
27. a 2 + 9a + 20. 36. 9a 2 -15a — 14.
28. a 2 -a-20. 37. 25 a 2 + 50 a + 21.
29. a*-*- 72. 38. c 4 - 13c 2 + 36.
30. c^ + iec + lS. 39. m 4 -29m 2 + 100.
31. m 2 + 8m + 16. 40. a 2 + 16 a? + 64.
32. tf-12y + ll: 41. ar-10s + 2&.
42. (a + 3)(a-3)(a + 5)(a-5) = ?
43. a 2 + 3a + 2. 45. a 2 + 2a + 2.
44. a 2 -3a + 2. 46. a 2 -2a + 2.
Polynomials having a Factor Common to Each Term
156. Type IV. Multiply 3a + ib + 2cby 5 x.
3a + 46 + 2c
5x
15az + 20&x + 10cx
Each term of the product contains the common factor 5 x. This is a
very important type of factoring. The first step to take in all examples
is to see if the example belongs to Type IV.
Ex. 1. Factor 4 a 2 x - 16 b 2 x.
Each term contains the factor 4 x.
128 MATHEMATICS
Divide by 4 x. 4 x)4 a*x - 16 6 a x
a 2 - 4 d 2
The divisor is one factor, the quotient the other factor, or
4 a 2 x - 16 b*x = 4 x(a 2 - 4 6 2 ).
But, a 3 — 4 6 2 can be factored by Type I.
Hence, 4 a^x — 16 6 2 x = 4 x(a + 2 b) (a — 2 6),
This is similar to factoring such a number as 105*
We remove a factor 3, then have
105 = 3(35).
Factoring 35,
105 = 3 • 5 • 7.
Ex. 2. Factor (2 a 4- &)»* - (2 a + 6)8 a + (2 a + 6)15.
2o-f 6 )(2q4-6)x 2 - (2 g 4- 6)8x4- (2 a 4- 6)15
a;2 _ 8 x + 16
Use Type III on the quotient. Then,
(2 a + 6)x 2 - (2 a + 6)8 x 4- (2 a + 6)15 = (2 a + 6)(x - 8;(x - 6).
EXERCISE 47
Factor the following :
(Be sure that you cannot still further factor your result.
Check each answer.)
1. 3^4- 21 x. 4. a 3 -ab\
2. a?b + ab\ 5. 5 ar>- 20.
3. 5 a + 25. 6. (a 2 -&y-(a 2 -6y
7. c*(2 m 4- 5) - c(2 m 4- S) - 12(2 m 4- 5).
8. (2c + 7)4c 2 4-(2c + 7)20c + (2c + 7)21.
9. ir(a? -}- 3) + y(x 4- 3). 10. ax + bx + ay + by.
Factor the first two terms and the last two terms separately.
x(a + b)+y(a + b).
The example is now like example 9.
21. a? — 3 x 2 — 3 x 4- 9. (x 2 is a factor of the first two terms
and —3a factor of the last two terms.')
PRODUCTS AND FACTORS 129
12. Factor (a + b)a 2 + (a + 6)2 ab + (a + b)&.
Compare example 9, exercise 27.
13. ^(2 x - 1) - x(2 x - 1) - 12(2 x - 1).
What Types did you use?
14. a\a 2 - b 2 ) - b\a 2 - b 2 ).
15. (2m + 5)m 2 + (2m + 5)8ra + (2m + 5)16.
16. (2m + 5)m 2 -(2m + 5)17m + (2m + 5)16.
17. (2m + 5)m 2 -(2m + 5)15m-(2m + 5)16.
la 25 ^(2 a; + 5)- 20 a(2 a; + 5) +4(2 a +5).
19. 4m 8 + 12m 2 + 36m. 20. 4m 3 + 24m 2 + 36m.
21. Is a 2 + 9 a + 25 a square? Why ?
22. <c 8 -12a; 4 + 36ar>.
23. (2<B + l)4a 2 +(2a; + l)4a; + 2a;+l.
24. (a 2 -a-72)(2a 2 -18). 2a ^ + 3 a^-Qx-lS.
25. (a + 3) 2 -(a:-3) 2 . 29. 8 ar>+ 12 a 2 - 10 a - 15.
26. (aj + 3) 2 aJ 4 -(a; + 3) 2 81. 30. a^ + 4^- 16 x- 64.
27. ^ + 3 a 2 - 9 x -27. 31. ra 3 -4m + 7m 2 -28.
32. 12 a 3 + 8 a 2 b - 27 ab 2 - 18 6 8 .
Trinomials of the Form ax 2 + bx + c
157. Type V. Find the product of 2 x + 5 and 3 * + 7.
3a; -f 7
2.3.se 2 + 3.5.z+2.7-z + 6.7
= 2 • 3 • x* + (3 • 6 + 2 • 7)x + 5 • 7
Notice that if the first and third terms of this trinorainal are combined
by multiplication, the product 2 • 3 • 5 • 7 • x 1 *, comprises all the factors
which make up the middle term of the product. Note also that the middle
term, 29z, is the sum of the cross products, that is, tfca sum ot It T v&d.
8z-6.
130
MATHEMATICS
Reversing this multiplication process we may factor trinominals of this
type.
Ex. Factor 60*+ 29 a; + 35.
1. Find the product of the first and third terms, or 210 x a .
2. Factor 210 x 2 so that the sum of the factors is the middle term 29 x ;
these factors are 14 x and 15 x.
3. Write the trinominal in the form of a quadri nominal, using 14 x and
16a for the middle terms :
Gx 2 + 29 a; + 35 = 6x 2 + 14s + 15a; + 35.
4. Now use Type IV.
2 x(3x + 7) + 5(3x + 7) = (3x + 7)(2x + 5).
Hence, 6 x 2 + 29a; + 35 = (3x + 7) (2 a; + 5).
It does not matter whether 14 x or the 15 x is connected with 6 x 2 .
Find by inspection :
1. (2» + 7)(3a; + 4).
2. (5x + 2)(2x + 4).
3. (6<c + 5)(2ic-3).
4. (5x + 3)(3x-4).
5. (2a>-7)(3a;-4).
6. (2<B-7)(3a; + 4).
7. (2<c + 7)(3a;-4).
8. (5a + 2)(2a-4).
9. (7a? + 3)(5a;-8).
Factor the following : 19. 6 x 2 + 23 x + 20.
6 x 2 - 20 = 120 x 2 .
120x 2 = 16x8x.
15x + 8x = 23x.
Then, 6 x 2 + 23 x + 20 = 6 x 2 + 15 x + 8 x + 20.
= 3x(2x + 5) + 4(2x + 5).
= (2x4-5X^x^4^.
EXERCISE
48
10.
(10 x -18) (7 x + 15).
11.
(8a;+17)(7a>-6).
12.
(4 x + 15)(9 x - 14).
13.
(6<c-21)(5a; + 18).
14.
(7x + 15)(7x + 15).
15.
(13s-l)(13a; + 2).
16.
(14s + 13)(13a; + 14)
17.
(14s-8)(14a; + 8).
18.
(3a;-75)(2a5-4).
PRODUCTS AND FACTORS 131
20. 6a? + 22x + 2Q. 29. 4 x 2 - 14 x - 98.
21. $x* + 22x + 15. 30. a^+lSa + Sl.
22. lOa^ + lOaj + a 31. 36 x 2 + 60 x + 25.
23. i23 2 -23s + 10. .32. 15^ + 14s- 16.
24. $x* + 2x-15. 33. 156 2 -14&-16.
25. 15ar ? + 23ic-28. 34. 15s 2 + 34a-16.
26. 15 ^ + 47 x + 88. 35. 10 a 2 - a - 24.
27. 56^-17 x-3. 36. 10z 2 -29a; + 10.
2a 4d 2 + 12d +9. 37. 14^ + 53^ + 14.
38. (2aj-3)8^ + (2a;-3)22a;+(2aj-3)15.
39. 8tf 4 + 2o 8 -15;c 2 . 40. 60^-115^ + 50^
41. (a 2 -9)4a 2 -(a 2 -9)4a + a 2 -9.
42. 4m 2 (2m + 5) + 12m(2m + 5) + 9(2m + 5).
43. 2 2 / 2 + y-28. 44. &c^-llc^^S6c.
45. 24m 3 + 14m 2 -3m.
Binomials having Both Terms Raised to the Same Power
158. Type VI. Such types arise from the following prod-
ucts : By actual multiplication
(a 2 + ab + ft 2 ) (a - b) = a 8 -
a 2 + ab + 6 2
a— 6
a 8 + a 2 6 4- «6 2
- a 2 6 - q6 2 - fr 8
a 8 -r- 6 8 .
Thatis, g ~ =g 8 + gft + 6*.
a — b
Note the form of the quotient : we shall speak of a as the
leading letter and b as the following lettfe?.
132 MATHEMATICS
1. The first term of the quotient contains the leading letter
raised to a power one less than its power in the dividend.
2. The power of the leading letter becomes less by one in
each succeeding term of the quotient.
3. The following letter appears in the second term of the
quotient and its power increases by one in each succeeding
term.
4. The signs of the quotient are all +.
a 4 - ft 4
Similarly, r- = a 8 + <?b + aft 2 + ft 8 ,
a— o
and *^z£ = a 4 + <fib + o*&* + aft 8 + & 4 .
a-
-b
a 5 -
-ft 6
a-
-ft
x*-
64
Ex. 1. ^ = a .2 + a ..4 + (4)».
a?— 4
=s 2 + 4 x + 16.
Here, a is x y
and b is 4.
Likewise,
and
^^L^-aft + ft*
a+ ft
? 5 ^ 5 = a 4 - (fib + a 2 ft 2 - aft» + ft*,
a + ft
Note that when the binomial divisior is all positive, the
terms of the quotient are alternately + and — . Do not divide
the sum of the same even powers by the sum of the roots.
EXERCISE 49
Factor the following:
1. x 3 -tf. 3. x 3 -8. 5. ar>- 27.
2. x* + tf. 4. ar*-f-8. 6. x* + 27.
7. 8 X s + 27. Here a = 2 a; and 6 = 3.
A 8a 3 — 27. *. laSrf-V-i.
PRODUCTS AND FACTORS 133
10. 125 a 8 -8. 17. a 8 + 243.
11. 64 m 3 + 27. 18. m 3 -216.
12. 64 m 3 -tf. 19. 5 3 + 512c 3 .
13. 64 m 3 + 27^. 20. 125 1/ 8 - 729 d\
14. 8 c 8 — 216 d\ 21. a? 6 + 27 (Regard a* as a cube.)
15. x 5 -32. 22. a! 6 + 64.
16. a 8 +32. 23. a 6 + 216.
Trinomials which may be referred to Type I by the addition
and subtraction of some number.
159. Type VII. x* + for 2 + c 2 . Usually in this type one
term is a fourth power, one is a square, and one is the product
of a square and some number.
Ex. Factor x 4 + 2 x 2 + 9.
If the middle term were 6 se 2 , this trinomial would belong to Type II.
Adding and subtracting 4 a; 2 , we have.
s* + 2x 2 + 9 = a* + 6 b 2 + 9 - 4z 2 .
= (x 2 + 3) 2 -(2a;) 2 .
By Type I, = (x 2 + 3 + 2x) (s2 + 3-2x).
EXERCISE 50
Factor the following :
1. x 4 + x 2 + 25. 6. a 4 - 39 a 2 b 2 + 49 & 4 .
2. 9# 4 + 83 2 + 4. 7 % a 4 + 4.
3. # 4 + 5a 2 + 49. 8. 42/ 4 -37# 2 + 9.
4. a 4 + 10 x 2 + 49. 9. 4^ + 82/ 2 + 9.
5. a? 4 -23s 2 + 49. 10. 25^-65^ + 10.
Fill in the term which will make each of the following a
trinomial square :
11. a 2 +4a + ? 12. 4 w? -V ^ m V*
134 MATHEMATICS
13. a 2 — 24 a?. 19. c 4 + 16.
14. c* + 4. 20. 9ra 2 -18m.
15. ? + 6a? + 9. 21. 25a 2 -S0ab.
16. a? + 8x + ? 22. 25a 2 -25a5.
17. x* + 5x + ? 23. 25a 2 -20a5.
18. c2^9c + ? 24. 25a 2 -18a&.
160. The seven types of factorable expressions should be
committed to memory. Before attempting to factor any ex-
pression, select the type to which it belongs.
Hints on Factoring
1. For all expressions :
First try Type IV.
2. Test binomials by Types I, IV, VI.
3. Test trinomials by Types II, III, V, VII.
4. Be sure that each factor will admit of no further' factor-
ing.
5. Several types may be needed to completely factor an
expression, e.g., example 13, exercise 47.
Types
I. a 2 -^ = (a + &)(a-&).
II. (a + b)(a + b) = a 2 + 2ab + tf.
(a-b)(a-b) = a 2 -2ab + lj*.
III. (x + a)(x+b) = x 2 + (a + b)x + ab.
IV. ad+bd + cd=d(a + b+c).
V. a^ + foc+c.
VI. a w -6 w .
VII. x 4 + bx* + <*.
PRODUCTS AND FACTORS 135
EXERCISE 51
Factor the following :
1. 2f-2y + l. 3. y*-2y-8.
2. tf-1. 4. y* + 2y-8.
5. 50ic 8 H-45« 2 -45a:.
6. 10ar*(2a-3) + 45aj(2s-3)-45(2a;-3).
7. (341) 2 -(339) 2 .
& One side of a square is 585'. This square is surrounded
by a concrete walk of uniform width, whose outside perimeter
is 2388'. Find the cost of the walk at 14^ per square foot.
9. One square is so placed within another that a space of
uniform width is between the sides of the outer and the inner
squares. The sides of the squares are 76" and 69", respectively.
How many square feet in the difference between the areas of
these squares?
10. The area of a square is 36 a 2 -f 84 x -f 49. Find one side.
How large is the square when x = — 4 ? If x were — 9, would
the area still be positive ? Could you substitute a value of x
that would make the area negative? Why?
11. The area of a square is 49 a 2 — 28 a + 4. Find one edge
when a = \.
12. Is a 2 + 10 a + 16 a trinomial square ? Why ?
13. The area of a triangle is 3 ^ + ^ * "~ 28 • Find the
base and altitude. If x is 2 and the triangle is isosceles, con-
struct the triangle. Is more than one such triangle possible ?
Why?
14. The area of a right triangle is 5 x 2 + 4 x — 12. Construct
the triangle when x = 2.
15. Find by inspection (2x + 3)(2x - S")(x "V ^ x "~ ^V
18. a 4 -81 = ?
136 • MATHEMATICS
16. Factor <xt - 25 x 2 + 144. 22. (4<e + i) 2 =?
17. a 4 -29a 2 + 100 = ? 23. Factor 4 x 2 + <e + J.
V 4 2x + lJ
19. (2m + 3) 2 -36 = ?
20. (2m-5) 2 -25m 2 = ? ^ 5 a? + l/ *
21. (2m-5) 2 -(2m + 5) 2 =? 26. Factor 5 a 2 + 11 a; - 36.
27. The area of a rectangle is a 2 -f 7 x + 12. Find its dimen-
sions when x = 4".
28. Find the dimensions of a rectangle whose area is
4o 2 -169y 2 .
29. Find the dimensions of a rectangle whose area is
225 a 2 - 289 b 2 .
30. Find the dimensions of a rectangle whose area is
x 2 -f-(3 m + 4 a) ar + 12 ma.
What are its dimensions when x = 4", m = 1", a = 1" ? How
does this rectangle compare with that in example 27 ?
31. Compare the adjacent sides in a rectangle whose area is
25x 2 -40xy + 16tf.
32. Factor 9 a* + 42 xcd + 49 cW.
33. Factor 4 c 2 - 256 c 4 d 2 m*.
34. 15ar J + 29a;-14.
In working examples under Type V, it is not necessary to
carry out all the steps indicated in § 157.
(15 «■)(- 14) = - 210 x 2 ,
-210x 2 = (35x)(-6x).
Connect either of these factors witti 15 a? \ fax exas&^le,
PRODUCTS AND FACTORS 137
15a? + 35a>.
Factor this expression, 5 x(S x -f 7).
Then, (3 x -f 7) is one factor of 15 x* + 29 x — 14, 5 x is the
first term of the other factor. The second term of the second
factor is formed by dividing — 14 by -f 7.
The written work would appear as follows :
Factor 15 x 2 + 29 x - 14.
(15 a 2 ) (-14) = -210 a 2 ,
-210x 2 = (35x)(-6x),
15 a? + 35 x = (3 x + 7)(5 x), .
15 a* + 29 »- 14=(3 x + 7) (5 a>- 2).
Factor :
35. 6^ + 14y-12. 42. 96^-24^-12.
36. 10m 2 + m-21. 43. 36 x 2 - 12 x - 120.
37. 40c 2 + 7c-3 = 0. 44. 483 2 + 50a + 2.
3a 6a^ + a?-126 = 0. 45. 31 a 2 - 151 a - 20.
39. 1032 + 51 OC + 56C 2 . 46. 48 a 2 + 128 a - 48.
40. 36 x 2 - 181 x + 225 = 0. 47. 25a 2 + 95a-20.
41. 30 x* - 27 a? - 21. 48. 64m 2 -40m + 4.
Solutions by Factoring
161. If the product of two or more finite numbers is zero, at
least one of the numbers is zero.
That is, if (x — 3) (x — 4) = 0, either x — 3 or x — 4 must be
zero. If a — 3 = 0, x = 3. If # — 4 = 0, a? = 4.
Such conditions give a method for solving equations which
are higher than the first degree.
Ex. Solve a 2 -4 a; = 21.
Transposing so that the second member of the equation is 0, we have,
138 MATHEMATICS
Factoring by Type III,
(z-7)(a; + 3) = 0.
Placing the first factor equal to zero, we have,
x = 7.
Placing the second factor equal to zero,
2C = -3.
These should be the roots of the original equation.
Checking for x = 7,
72-4.7-21 = 0,
49-28-21 =0.
For x = - 3, (- 3) 2 - 4(- 8) - 21 = 0.
9+12-21 = 0.
Both roots satisfy the equation.
EXERCISE 52
Find value :
1. 24(» + 5)(a-3)(a; + l), whens = 0.
2. 16 (x - 1) (x + 2) (x + 8), when x = 1.
3. 1625 (x + 3) (x + 5) (x - 16), when x = - 5.
Solve by factoring. (Check each root.)
4. x 2 -7x + 12 = 0. 11. v 2 -8v = -16.
5. 2/ 2 + 5y + 6 = 0. 12. aP-x = 12.
6. 2/ 2 + 7# + 6 = 0. 13- m 2 — 4m = 0.
7. a? -81 = 0. 14. 48a 2 -12s = 0.
a ar , -9ar J -9aj + 81 = 0. 15. y 2 -ll y-102 = 0.
9. 4» 2 + 12a; = -9. -16. 9a 2 -30z + 25 = 0.
10. tf-tf-9y + 9 = 0. 17. 16 3^ + 42 aj = ~5.
18. (2a + 7)a 2 -(2a + 7)4a+(2a + 7)4 = 0.
19. (^-16)4 x 2 -(x 2 -16)2Sx + (a 2 -16) 49 = 0.
20. (25 z 2 - 225) (49 z 2 - 289) = 0.
21. How many roots to a first degree equation? second
degree? third degree? nth degree?
PRODUCTS AND FACTORS 139
Supplemental Applied Mathematics
1. A 10-inch steel pipe is 10.19" inside diameter, 10.75"
outside diameter. Find weight per lineal foot.
2. A 9-inch pipe is 8.937" inside diameter and 9.625" out-
side diameter. Find weight per lineal foot.
3. A steam pipe is 12.750" outside diameter and is made of
steel f " thick. Find the inside diameter.
4. The inside diameter of a 2-inch pipe is 2.067". If water
is forced through this pipe at the rate of 10' per second, how
many gallons can the pipe deliver per hour ?
5. A 6-inch main, inside diameter 6.065", in the same system
as the 2-inch main above would deliver how many gallons per
hour? Could you use the result obtained in example 4 in
solving this problem, and have a fairly accurate result ?
6. A pump delivers 23.5 gallons of water per stroke and is
set for 16 strokes per minute. What weight of water is de-
livered per hour ? (Water weighs 62£ pounds per cubic foot.)
7. A technical school needs 75,000 sheets of paper
8J" x 10£". The stock is 17" x 22", 16 pounds to the ream of
500 sheets, and costs 6f ^ per pound. How many reams must
be ordered and what is the cost ? (This is a problem in mental
arithmetic.)
8. 30,000 cards 5" x 4" are to be cut from stock 17" x 20|".
Weight 30 pounds to ream, price 12^ per pound. How much
stock was bought ? What did it cost ? Does it make any dif-
ference which way the stock is cut ?
9. In a fuel test 100 pounds of coke was found to contain
6.01 pounds of ash and .583 pounds of sulphur. The balance
was carbon. What per cent of carbon was there ?
10. Air is .001276 times as heavy as water. What is the
weight of the air in your classroom?
140 MATHEMATICS
U. If a body falls freely in space, the distance fallen is equal
to %g times the square of the time in seconds, where g is the
force of gravity (32.15 feet).
The equation for this law is usually written
s = ^ gfi, where s is the distance in feet.
A stone dropped into a canon is seen to strike the water at
the bottom of the canon in 8 seconds. How deep is the canon ?
12. A stone dropped over a precipice fell for 10 seconds be-
fore striking the ground. How far did it fall ?
13. An aeroplane was sailing 1000 feet above the ground
when one of the passengers dropped a handbag overboard.
How long did it take the handbag to reach the ground ?
14. 56 hurdles 5 feet long just reach across a field. How
many hurdles 4 feet long would be needed ?
15. For bronze bearings the Pennsylvania Railroad uses the
following alloy, 77 % copper, 8 % tin, 1 5 % lead. How many
pounds of copper, tin and lead are used in making 900 pounds
of bearings ?
16. In one lot of 402 castings, 24 were spoiled, in a second
lot of 500 castings, 38 were spoiled, in a third, lot of 321, 22
were spoiled. In which lot was the largest percentage of loss
due to spoiled castings ?
17. In testing our shop drive, 9.24 horse-power went to the
lathes and .75 horse-power went to the grindstone. The motor
delivered 11.2 horse-power. What per cent of the power went
to the belting and shafts ?
18. My competitor and I handle hardware. For the same
set of articles my competitors' prices are $ 3, $ 3.30, $ 3.55 and
$3.70. His trade discounts are 25%, 7£, 5 and 2. My list
prices for the same set are $6.10, $6.70, $7.20 and $7.50 and
my trade discount is 60 %>, 7£, 7£, 5 and 2. In making a bid
how do the net prices compare ?
PRODUCTS AND FACTORS 141
19. Three men can set up a line shaft in 8 days. How many
men can set up the same shaft in 3 days ?
20. Fourteen men are at work installing the machinery in a
shop. They work for 8 days and finish half the work. The
work must be completed in 5 more days. How many men
must be added ?
21. 200 men were completing the work on the Technical
High School. The job had to be completed Oct. 5. On Oct. 1,
the contractor found that there were still 500 days' work to be
done. How many men could he lay off and finish the job on
time?
22. One cup ground coffee makes 6 cups boiled, £ cup
boiled coffee serves 1 person. How many level tablespoonfuls
of ground coffee should be used for each person ?
23. If coffee costs 35^ per pound and there are 4 J cups per
pound, find cost of enough beverage for one person.
24. For filtered coffee $ cup is used with 5 cups water. Find
difference in cost of enough boiled and filtered coffee to serve
6 portions.
25. To clear 1 quart coffee, £ white of egg or several egg
shells may be used. It takes 11 yolks or 9 whites of eggs to
measure 1 cup. Assuming that the yolks can be used for other
purposes, eggs selling at 30^ per dozen, and 1 quart coffee
used daily, how much can be saved in a month by using egg
shells to clear the coffee ?
26. Tea made from Ceylon tea leaves contained 8.6 % tannic
acid after five minutes' infusion; 10.88% tannic acid after
thirty minutes' infusion. Find the difference in the quantity of
tannic acid extracted from tea leaves steeped for 5 and 10
minutes during a month, if % pint tea is used daily and 1 pint
of the beverage weighs 1 pound.
27. A tea contained 6.8% tannic acid after five minutes'
infusion and 16.3 % tannic acid after forty minutes' infusion.
Find the difference in the quantity oi ta&ma fcfc\& ^i&ra&j&t
142 MATHEMATICS
from the tea leaves after steeping 5 and 40 minutes during
a month if f pint tea is used daily.
28. Green tea leaves contain 10.64 % tannic acid ; black tea
leaves contain 4.89 % tannic acid ; 1 cup tea leaves weighs 2
ounces. If 1 teaspoonf ul tea leaves is used in making a cup of
tea each day, find the difference in the quantities of tannic
acid extracted from black and green tea during a month. One
cup tea leaves measures 48 teaspoonfuls.
CHAPTER IX
Fractions
162. A fraction is an indicated division. It is written in the
form -, the number above the line being the numerator • or
dividend, the number below the line being the denominator or
divisor.
A fraction may be positive or negative (see Chapter III), the
sign + indicating that the quotient is to be added, the sign —
indicating that the quotient is to be subtracted.
Thus, — ^- means that the quotient arising from — 4 -=- 2
must be subtracted, the result is + 2.
We see that three signs are involved in every fraction. The
sign of the numerator, the sign of the denominator, and the
sign of the fraction. Any two of these signs may be changed
without changing the value of the fraction.
'4 _ 4 __4 4
Thus, +-= ± = =±=-±-.
V 2 2-2-2
And «-3 = _ q-3 = — a + S _ _ -a + 3 _ _ S-a
a + 6 -a-Q -<*-6 a + 6 a + 6"
Such changes often simplify operations with fractions.
Principles of Fractions
163. The following principles govern operations in fractions :
1. Multiplying the- numerator of a fraction by a number multi-
plies the fraction.
IAS
144 MATHEMATICS
This depends on axiom 3, § 23,
{-*
Multiply both sides of the equation by some number, 5, we have,
a
2. Multiplying the denominator of a fraction by a number
divides tJie fraction.
The pupil may show that this principle depends on axiom 3.
3. Multiplying both numerator and denominator of a fraction
by the same number does not change its value. Why ? Axioms ?
4. Dividing the numerator of a fraction by a number divides
the fraction.
D
-- — Q. Then - = " . The pupil may explain use of axioms.
d a 5
5. Dividing tJie denominator multiplies the fraction.
The pupil may illustrate.
6. Dividing both numerator and denominator by the same
number does not change the value of the fraction. Explain.
Reduction
164. Principles 3 and 6 are involved in the reduction of
fractions, 3 in reduction to higher terms, 6 in reduction to
lower terms.
Ex. 1. Reduce f to higher terms.
6 4_20
6*4 24*
Ex. 2. Reduce f to 48ths.
6=2 -3; 48 =2*. 3
2-3 )2**3
2»
FRACTIONS 145
Hence, multiplying both numerator and denominator by 2 s or 8, we have,
5 8_40
6*8 48*
o
Ex. 3. Reduce to a fraction whose denominator is
a + 6
a 2 -a-42.
a 2 _ a __ 42 = ( a + 6) (a - 7) .
a + 6)(q + 6)( a-7)
a - 7
Hence, multiplying both numerator and denominator by a — 7, we have,
<* - 3 _ (a - 3) (a - 7 ) _ a 2 - 10 a + 21
a + 6 (a + 6)(a-7) a 2 -a-42 \
EXERCISE 53
1. Reduce to 144ths : ^, ^, ^, ^.
2. Reduce to 512ths : ^, -^, ^.
3. Reduce to 19ths : %, \, \.
4. Reduce to fractions whose denominator is a 2 — a — 12.
q + 4 a-7
a-f-3' a-4'
5. Reduce to (4 x 2 + 12 a? + 9)ths : V*"~\ .
6. Reduceto(4« 2 -12aj+9)ths: ^ a? + ^ .
7. Reduce to (a? — 5) (x + 5) (a; + l)ths :
2 4aj-l 2a> + 7 1
x*-25' a^ + 6» + 5' a^-^aj-S' a> + l
8. Reduceto(4aj-l)(2aj + 5)(3aj-l)ths:
-4 3 a? + 2 8a? + 5 -(a? + 2)
Sa^+lSa-S' 12^-7^+1' 6^+133-5' 12rf-7aH-l'
9. Reduce to (4 x 2 — 9) ths :
Ss + 2 . 2 ^_±i. 5a? + 2 («i62\
2a? + 3' 2aj + 3' -2x^-S v %)
146 MATHEMATICS
10. Reduce to 14 (a 2 — a — 72)ths:
8s-l . -5 11
7(»-9)' 2 (a + 8)' 14*
165. Reduction to lower terms.
Ex. 1. Reduce $ f to lower terms.
36 _ 2 a . 3 a
64 2 • 3» '
By inspection we see that 2 • 3* are factors common to both numerator
and denominator. Dividing both numerator and denominator by 2 • 3 2
(principle 6), we have
2 2 . 3 2 2 *
— = 2 • 3 2
54
2 • 3 8 3
_2 fm *J O
Ex. 2. Reduce — — — - to lower terms.
a 2 -9a + 20
a 2 - a -12 = (g-4)(a + 8) ,
a 2 -9a + 20 ( a -4)(a-5)'
Dividing both numerator and denominator by the common factor a— 4,
we have
(q-4)(g + 3) _ q + 3
(a — 4) (a — 6) a — 6
« 2 -«- 12 =(a-4)
a 2 - 9 a + 20
EXERCISE 54
Reduce to lower terms:
11. «'- 9 . 13. «±i. 15. ^ + 3^-10
95
TTff*
54
a 2__6a + 9 m 2 + l 2z 2 + 3z-U
m 2 + 2m + l 14 m 4 -64 8c?-27c P
m 2_l ' ' m 4 -16m 2 + 64' ' 4c 2 -9d 2 "
4c 2 + 4cd-15d 2 10 8c 3 -27d 3
A7. - — i 1 • A8.
8c s -27d 3 ' 12 c 8 + 18 c 2 d + 27 cd 2
* The bar / as here used indicates division oi tooth numerator and denomina-
tor of the fraction.
FRACTIONS 147
12 <* + 12 <?d-±5cd 2 9y 2 + 18y3-16s 2
12c 3 +48c 2 d + 45cd 2 ' * 18y 3 + 36y 2 3-32y* 2 '
9y 2 -12ys + 4s 2
9^-4» 2
22.
27y 8 -8g 3
(18^-12^X9^ + 6^ + 4^)'
23 e^-lSyg + eg 2
' 32/ 2 + 19yz-14s a
24 (y + 7g)9y 2 -(y + 7g)12yg+(y + 7z)42? 8
3^ + 19^-14^
(3^-2^)9^ + 18 ys(3y-2s)-(3y-2s)16s 2
(27^-80aO(3y-2*) f
Multiples
166. A common multiple of two or more numbers is a mul-
tiple of each of them, e.g., 48 is a common multiple of 12 and
16. A common multiple must contain all the factors of the
numbers involved.
Ex. 1. Find a common multiple of 18 and 24.
18 = 2 • 3 2 .
24 = 2» . 3.
The different factors concerned are 2 and 3.
The common multiple of these numbers must be made up of the prod-
uct of 2 at least three times as a factor and 3 at least twice as a factor.
That is, a number cannot contain 18 an integral number of times unless
it has as factors 2 • 3 2 . It cannot contain 24 an integral number of times
unless it has as factors 2 8 • 3. To contain both 18 and 24 it must at least
have the factors 2 s • 3 2 . The common multiples of 18 and 24 are, therefore,
2*.3 2 , 2*-3 2 , 2«. 3 8 , 2*. 38, 2 8 . 3 2 • 5, etc.,
or 72, 144, 216, 312, 360, etc.
167. The lowest common multiple contains all the factors of
the numbers involved the least number oi \»m&fc.
148 MATHEMATICS
To find the 1. c. m., form a product of all the different factors
of the given numbers, and give to each factor the highest expo-
nent found in any of the numbers.
Ex. 1. Find the 1. c. m. of 45, 48, 54.
45 = 3 2 . 5.
64 = 2 • 3 8 .
48 = 2* • 3.
The 1. c. m. is, therefore, 2* • 3 2 • 5.
Ex.2. Find the 1. c. m. of a 2 - 9, a 2 -6a + 9, a 2 + 3a-18.
a 2 -9 = (a+8)(a-3).
a 2 -6a + 9 = (a-3) 2 .
a 2 + 3 a - 18 = (a - 3)(a + 6).
The 1. c. m. is (a - 8) 2 (a + 3) (a + 6).
How many times will this 1. c. m. contain a 2 + 3 a — 18 ?
Ex. 3. Find the l.c.m. of a — 1, x* + 9, a — 5.
These numbers are all prime, their 1. c. m. is their product, or
(3-l)(a; 2 + 9)(x-6).
EXERCISE 55
Find 1. c. m. :
1. n s + 2n 2 , w 8 -4n. 3. 72, 48, 27.
2. a 2 -9, a 2 - a- 12. 4. 120, 18, 30.
5. 20, a? 2 - 16, 3ar J -6a;-24.
6. 15, 3a?- 13 a- 10, a?-25, 12.
7. 4a»-l, 12a? 2 + 12a? + 3, 4 a,- 2 - 4 x + 1.
8. ^-27, 2 2 + 3z + 9, ^-9.
9. m 2 + 3m-10, 2m 2 + 7m-15, 2m 2 -7m + 6.
10. <c + 3, x — 3, a? 2 — 6.
Addition
188. Fractions, like other numbers, must be of the same de-
nomination before they can be added. To Tfedxuife ix^vsaa^
FRACTIONS 149
the. same denomination; the lowest common multiple of their
denominators must be found, and principle 3 (§ 163) employed.
Ex. 1. Add ^ T \, ^.
By § 167, the 1. c. m. of 24, 18, 16 is 2* . 3*.
In such work as reducing to 1. c. d., always divide by factors.
and
2*
.32-5
r24 =
:2-3.
Multiplying
7 and 24 by 2 -
3, we have
7
7-
23
42
24
24
.23
144
Similarly,
6 .
18"
6-
18
2«
.28
40
144'
i
3
3-
32
27
16
16
• 32
144
Then, I + A + A = i2 + iO + ^I = 10?.
24 18 16 144 144 144 144
Ex. 2. Find the sum of
a + 2 a 2a-l
a 2 -25 a 2 -10a + 25 3 (a 2 -2a- 15)'
a 2 -26=(a + 6)(a-5).
a 2 - Ida + 26 = (a - 6) 2 .
3 (a 2 - 2a - 16) = 3 (a - 6) (a + 3).
1. c. m. = 3 (a 4- 6) (a + 3) (a - 6) 2 .
3 (a + 6) (a + 3) (a - 6) 2 -4- (a 2 - 25) = 3 (a + 3) (a - 6).
ThBn q + 2 _ 3 (a + 2) (a + 3) (a - 5)
' a 2 -26 3 (a + 5) (a + 3) (a -6)2*
3 (a + 6) (a + 3) (a - 6) 2 -*- (a 2 - 10a + 26) = 3 (a + 6) (a + 3),
and a = 3a (a + 6) (a + 3)
a*- 10a + 26 3 (a + 6) (a + 3) (a- 6) 2
and 2q-l = (2q-l)(q + 5)(q-5)
3 (a 2 - 2q - 16) 3 (q + 6) (q + 3) (a - 6) 2
Hence, 4±± + a 2a ~ 1
a 2 -25 a 2 -10a + 25 Ktf-^a-^
150 MATHEMATICS
_ 3(q+2) (g + 3)(g~5) 3g (q + 5)(q 4 3)
3 (q + 5) (a + 3) (a - 5)2 + 3 (a + 5) (a + 3) (a - 5)3
(2a -1) (a + 5) (a -5)
3 (a + 6) (a + 3) (a - 5)*
_ 3 « 8 - 57 a - 90 + 3 a 8 -f 24 qg + 45 q — 2 q* 4 fl 2 4- 50 q - 25
3 (a 4 6) (q + 3) (q - 6)2
(Note the change of sign in the last numerator. Why f Always watch
for such negative numerators).
_ 4q* + 25q2 + 38q-115
3(q + 5)(q + 3)(q-6)2'
Ex. 3. Find the sum of —L_ ?_ + 3
x — 2 x — 3 a; — 4
1. c. m. = (x - 2) (x - 3) (x - 4).
Then, -J ?- + 8
x-2 x-S x-4
= 1 (x - 3) (x - 4) - 2 (s - 2) (x - 4) -f 3 (x - 2) (x - 3)
(x-2)(x-3)(x-4)
_ a 2 „ 7 g + 12 - 2 a* + 12a; - 16 + 3a* _ I5x + 6
(s_2) (x-3)(x-4)
= 2x2-10x4-2
(x-2)(x-3)(x-4)'
EXERCISE 56
Find the following sums :
1 + -±-=- 5. _«_ + b
x + 2 x + 5 a + b a — b
9 1 1 fi C + 1 C ~ 1
2. — — • O. •
x + 2 x + 5 c — 4 c4-4
3. ^i + «±l. 7. -»- + 8
ar"-4 x + 2 <?-25 <? + 5c
. x + 2 x + Z _ mm .
' x*-9 + x*-6x+9' ' m + 1 »-l
9. *_ + 5 5
*'—v—12x*—16 ar l + lx + 12
FRACTIONS 151
10 . l +£=*£. u. l- ^-W . 12. £±»-«Il»,
(a+6) 2 (a + 6) 2 a- b a + b
13 5(o?-fy) 6(x-y)
a> 2 + 4;cy + 3y 2 x 2 + 2xy — 3y 2
14 1 * I 6 16 ?^±i - JLtJL 4- 3 "
* 2a+3 2a-3~ r 4a 2 +9 ' 2a-l l-4a 2 " r '
15.
2a? + l 4 + 5a?
2x-l l-2x
17 a^ + l 4a; _ a? 8 + 3s 2 + a; + 3 + 4a?--4a;
" s"-l * + 3 (rf-i)(aj + 3)
^ gE + TaE-Sog + S ,
a^ + 3a?-aj-3 '
If the degree of the numerator is equal to or greater than that of the
denominator, the fraction is improper, and may be reduced, as in arithme-
tic, by dividing the numerator by the denominator. Then
9E?+8s*-x — 3 tf+Stfi-x-S'
Always reduce the result to its simplest form :
ia l+-g-+ 4 . 20. £±& + ^&.
i9 a '+* 1 a -s 21 &y- Sc 1 5 .v+ 3c .
' a? + 5 a + 2* ' 25y 2 -9' r 5y-3
5 y — • 3 c 5 y -f 3 c
25y 2 -9c 2 5y-3c'
2c 2 + 7c-4 ^^H-Sc-S
2c 2 + llc-6 2c 2 + 5c-3'
23.
24 3m 2 + 13m-10 3m 2 + 17m + 10
" 3m 2 + 17m + 10 3m 2 + 13m-10
a 2_ a _56 a 2 -25
25.
a 2 + 14a + 49 a 2 -2a-35
M 2a + 2b . 2 a - 2 6 , 8a6
z& 1 r -- r
a-b a + b a 2 — tf
152 MATHEMATICS
a-\-b a — b 4 aft
27.
a—b a-\-b a 2 — b
28 3c + 5 9c* + 25 g-2 , s + 2 3*-4
' 3c-5 9c 2 -25 ' a? + 2 x-2 x* + ±
30. . V 8 - . * + 5 + *~ 7
aj2_2a?-35 x*-10x + 21 x* + 2x-15
31. -J C —+ *
c _2 c>-4 c*-8
Express as a single fraction :
32. 4 + £=|. 33. 2^ + 2- 34. 4+|-l.
sc + 5 a 2 + 9 a 2 — 9
35. 2 + — — H - 38. i-i- + -.
»» - 25 6 »■ - 125 75 81 *
*■ #*+«-** ^ 5a , 14s ,
39. — — -+- — — X.
37 4i 8i i 76 95
169. To reduce a fraction to a whole or mixed number
principle 6 must be employed. We choose the denominator
of the fraction as the number by which the numerator and
denominator must be divided.
Ex. Reduce „"~ — — — x ~~ to a whole or mixed number.
2o? 2 -3a; + 7
Dividing both numerator and denominator by 2 x 2 — 3 x + 7,
wehave 5«+J§_ «« + *"
2 4 4(2x 2 -3x + 7)
This result is a mixed number, and the first two terms are integral.
170. An algebraic expression is integral if its denominator
is numerical. That is, if its denominator is 1, 2, 3, 4, etc.
In the result of the example in § 169, f x and -^ are in-
tegral algebraic expressions, though not integral arithmetic
expressions.
FRACTIONS 153
EXERCISE 57
Reduce to whole or mixed numbers :
4 x + 1 6 x* + 5x*-lx + §
Ax-1 ' x 2 ^Sx-2
x* + 5x + 6 7 8^ + 64^
aj + 2 * " 2x + 4y
- s* + 5s4-6 fl s 4 4-81
3-3 a + 3
,. 7aj 2 -5a? + 12 ft aj + 1 , a?4-2 , * + 3
5.
10.
3a? + l a?4-2 a + 3 aj + 4
a^ + Sa^ + Sg + l . (Reduce each fraction sepa-
x 2 4- 2 a? + 1 rately, then add the results.)
x—1 x — 2
x-2 x-3'
(Reduce to mixed numbers, then combine.)
11 a + 4 a?4-5 12 a?4-2 s + 3 2a? + 5
a + 5 a?4-6' a? + 3 a + 4 a + 2 *
Multiplication
171. Multiplication of fractions involves principles 1 and 6.
Ex. 1. Multiply %$ by 5.
1 q IRK
_? . 5 = _2J_5. (Never use cancellation.)
25 26 v
18_^6 =6
25
Ex. 2. Multiply ^| by 3.
18. 5 _ 18
25 5
18 3 _ 18 » 3 _ 54
25 * 25 25
Ex. 3. Multiply a + 2 by a + 4.
a 2 — a — 12
a +2 , , i N_ (a4 2)(q4 4 )_ a 2 + 6q4-8
(a-4)(a + 8)
154 MATHEMATICS
Ex. 4. Multiply . a + 2 by a - 4.
a 2 — a — 12
_a±2_ u (g - 4)= (« + 2)(a-4) =s a + 2 < (Principle60
Ex. 5. Multiply f by T V
5 A 54
- . 4 =
8 8
But our multiplier is ^ of 4. Hence our product is
Li + 15 or Li. = i. (Principles 2 and 6.)
8 8-15 6 v F '
EXEBCISB 58
Simplify, using factoring :
196 289 361
- x ~ 1 (x-2). 5. a2 ~ 5a + 6 .(a 2 -9).
^-4 v ' a 2 -9
m 2 + 6m 4- 8
m 2 + 7 m + 10
(m 2 + 5m).
7 . M . M . g+86. 16y-8ft + l
46-1 36 + 24
a fe^' ^°iS - ii i8 « 2 * i2av i5 -
g c + 5 c^-lOcH-^ 5a» 98 6* 75c 8
' c-8 ^-25 " 7bc' 125ac' 2$ab'
m 2 + 5 ?n, + 6 m 2 + 7 ra + 10 m + 1
13.
14.
15.
ra + 5 m 2 + 4m-f-4 m 2 + 4 m + 3
10c 2 - 21c -10 3(^-80-3 4c 2 + llc + 6
12c» + 13c + 3 ■ ' 2c 2 -c-10 ' 5c?-13c-6
/c s +7c + 6 \f ± _ 12c + 17 V
(c 2 -3c-'10j\ c 2 + 8c+12J
FRACTIONS 155
__ ig' + igy + y* x + 9y
" ^-81^ ' (x-y)(x>+xy + f)'
19 . (i + *±^V (H . «L_A
^ s + 5 g + 3
a + 3 a + 5
V a? + 6x + 9 J
Division
172. Division depends on principles 2 and 6.
Ex. 1. |-5-3.
9. 3 _P = 3 2 _3
5 " 6-3 63 5 #
Ex. 2. Divide ^4 by a? - 1.
ar — 4
a;2__4 ^ ' (x + 2)(a;-2)(a;-l) a; 2 -4
Ex. 3. Divide | by ft.
7 ^21 = 7
8 8- 21
But our divisor is ^th of 21, hence our quotient is 12 times as large as
though the divisor were 21. Hence,
7 ■ 21 = ( 7 Wr 7 ' 12 ^ 7 - 22,3 ^ 1
8 ' 12 \8-2iy 8-21 2 8 3-7 2
1 7 • 2 2 • 3
What divisor was used to produce - from * * , ?
* 2 2 8 • 3 . 7
Therefore, to divide a fraction by a fraction invert the terms
of the divisor and proceed as in m\iU\^\\^\i\OTi.
156 MATHEMATICS
EXERCISE 59
Simplify the following :
14 . -288(a-6 2 ) . SSja-b 2 ) 1
* 15 ' 289(a 2 -6) "*" 51 (a 2 - 6) 2 *
o *2? f_-m 7 a 2 -2 5 a 2 -8a + 15
105 "^ ; * a 2 -9 ' a 2 -6a + 9 '
3. -
102 /-17\ Q a 2 -17 a + 16 or *„.
A 50a 2 6 1506 2 ft a-& a + 6
33 c 3 44 c 4 a+6 a- 6
169(s+y) 2 . 13(s+y) *-27 . ** + 3z + 9
' 144(a-2/) 8 ' I8(x-yf ' z 2 + 9 ' 3^ + 273 '
10^ + 170 + 3 . 2^ + 170 + 21
" 2c 2 + 13c-7 ' 10c 2 -3c-l '
*a? + 2a;-24 1 s 2 -4qa? + 4a*
_ a 2 + 10a; + 21 , A 8os
12. -s -r-s: s^» 13.
a? 2 + 12 x + 36 * (a? + 2a 2 )
ar* + 3 x - 28 4 ax
14.
(*±!5±*-iVc.-*
1+ 1
xy _ » + V a? — V
a? y x + y a; — y
s~y ar + y ^ b
y x c 6 + c
17. —1 ■ 18. r + r^-'
x-y x+y 1 _6 6— c
x y c
19. Divide 6 - c b y c - 6. 20. ~ ^-ff^ g+JLr
ar + y 2 x — 2y
* Division written in this form ia called a com^\*xtm,\toa..
FRACTIONS 157
' (115y(x + y y * (115)*(x' + 2xy + tf)
22.
6 _a * &!_o 4- —
a 6 a 2 6 2
23. Transform 24. Transform
l- b - „ . 2c + l
#
<Lz| into _i. |J±5 into
a + b 1 b 2c-x 2c_ 1
a x
Fractional Equations
173. Equations involving fractions are solved by first reduc-
ing to integral equations by means of principles of § 167, and
Axiom 3, then solving as in exercise 8.
Ex.1. Solve ££±1 = 5-?.
X X
Thel. c. m. isx (§ 167).
Multiplying both sides of the equation by x (Ax. 3),
x
m=< 6 -i)
We have, 2 x + 7 = 5 x — 2,
or — 3 x = — 9,
x = 3. (Ax. 4.)
Check this root by substituting in the given or original equation.
2 - 3 + 7 = 5-g, or 15 = 5-1.
3 3' 3
Ex.2. Solve *Ill = ^! *
«-2 x-3 2
1. c. m. = (x-2)(x-3)2.
Multiplying by 1. c. m.
2x 2 -8x + 6 = 2x 2 -8x + 8-x 2 + 6x-6.
Transposing, x 2 — 6 x + 4 = 0.
Factoring by Type III, (x - 4) (x - 1) = 0.
Solving by § 161, x = 4 at \.
158 MATHEMATICS
4-1 4-2 1
Check for x = 4.
or
For x = 1,
4-2 4-3 2'
? = ?_!
2 12'
l-l = l-2 1
1-2 1-3 2*
Ex.3. -1-+ * 3a; - 6
05 — 5 a + 4 a 2 — a; — 20*
1. c. m. = (x — 5) (a; + 4).
Multiplying by 1. c. m.
x + 4+x-5 = 3x-6.
Transposing, — x = — 5.
x = 5.
Check: J-+ * 3-5-6
6-6 5 + 4 52-5-20
1 . 1 9
or - + - = -.
9
I?w$ iY is not allowable to divide by zero.
Hence 5 is not a root. The equation has no solution. It is hot an
equation of condition. It is simply a statement that two numbers are
equal, § 14, but the statement is not true.
EXERCISE 60
Solve and verify the following:
1. ^ = 8. 5. *g + l~S* + ±.
3 6. -J^+ 1 - 3 * + 2
2. ^- = 8. x-1 x-2 x*-Zx+2
2 x
„ 1 , 1 5x-9
7. T +
K 15 x-1 x-2 x 2 -3x+2
3. 5 = — .
x
2x . 3s + 4_6«-15
_ 2x-l 15
* ^ — — • (Unite the first and third fractions be-
fore clearing tne ec^aXXaii tf.txAjdtasQ&:^
x
FRACTIONS 159
2x 3a + 4 _ 6a? + 5
" 5 + 2a?-l 15 '
10. ? + £i! = 4a>-2a. Solve for a?.
2 3
11. ! + ! = £ + ?. Solve for a;,
a: a x a
12 4y-l 2y + 7 ^ 8y + 7
6 4t/-8 12
13. H^ = ?. Solve for*.
* — a 4
4 £ -f a _ £ — a _ £ — 3 a
t — a £ + a £ 2 — a 2 *
,_ 6a? + 14a? + 8 _ 4(3s--5)(a? + 4) ,
is. — ^— — -, ^ ' ~t~ a?.
3 6
a? a; a — b
16.
17.
a + 6 a — 6 (a + &) 2
15 ^ + 5 22 Sz
9z*_25 5-f-3» 5-32* .
^ 3m 2 -2m + 13m 2 + m-2 = -19m 2 + 3m + l
m + 3 m-3 m 2 -9
, ft m-f a , 2ra — a a , -
19. — ■ h = 2. Solve for m.
m — a 2m -fa
_ 4g + l 2z + 3 32-5 122-4
4SU. *— — ss •
3 5 10 15
^ 3fc + a 5fc-a 2fc + 3a ^ 7fc + 2a
2a 4a 6a 10a
Solve for k.
9Q 5z + 2cP 22 + 8a" 2 152-d 2 C1 ,
22 ' ~7* T^s—W Solvefor *-
23. -L. + -1-- 6 *-26 .
j.-5 y-6 ^-lly + 30
160 MATHEMATICS
m _5 m-6 m 2 -llm + 30 #
25 6fc 2 fc = 8(fc*-4)
* fc + 5 2-& A? + 3&-10*
11 1 1
26.
z— 2 2—4 2—3 2—5
(Combine the first two fractions, then the second two fractions before
clearing the equation of fractions.)
«„ cy + my 1 , 1
c c — m m
28.
eg — 3 d-4 = d-l ri-2
d_4 d_5 tf_2 d-3
„_ x + 1 . x~2 2x + 8
29. — ! 1 = ! — •
x + S a; — 4 # + 4
30. The denominator of a fraction is 5 more than the nu-
merator. If 9 is added to the numerator and 7 subtracted from
the denominator, the result is 1£^. Find the fraction.
31. Find three consecutive numbers which will satisfy these
conditions : if the smallest is multiplied by six and three sub-
tracted from the product, this product divided by the greatest
number will give a quotient of 5.
32. I have two proper fractions. The denominator of each
is one more than its numerator, the numerator of the first, the
numerator of the second, and the denominator of the second
are consecutive numbers, and when the greater fraction is sub-
tracted from the lesser the quotient is — ^. Find the frac-
tions.
33. Divide 48 into two parts, such that the fraction formed
by these parts is ^.
34. Two Ohio cities are 120 miles apart. Two trains run-
ning between these cities have a difference in rate of 5 miles
per hour, and the difference in time it takes them to make the
run is 20 minutes. Find the rate oi eaek tavou
FRACTIONS 161
35. A local train loses six of its passengers at the first stop,
one third of the remainder at the second stop, one half of the
remainder at the third stop, and the 30 who then remain ride
to the end of the line. How many passengers on the train
when it started ?
36. In example 35, the first station is £ as far out as the
second, the third is 36 miles beyond the second, the fourth 49
miles beyond the third, and the length of the run is 117 miles.
At 2^ per mile per passenger, how much did the railway
company receive?
37. "A number exceeds the sum of its one third, one fourth,
and one fifth by 13. What is the number ?
38. An oil tank can be filled by one pump in 8 hours, and
by a second pump in 14 hours. How long does it take to fill
the tank when both pumps are working ?
Let x = the time when both pumps are working.
Then - is the amount of work both pumps do in 1 hour.
x
- is the amount of work the first pump does in 1 hour, and
o
-— is the amount the second pump does in 1 hour.
14
Then i + -L = l. Solve.
8 14 x
39. An oil tank 20' in diameter can be filled by one pump
in 177£ hours, and by a second pump in 64 hours. How long
does it take to fill the tank when both pumps are working ?
40. In example 39, the cylinder of the larger pump is 10"
in diameter, the stroke of the piston is 12", and there are 15
strokes of the pump per minute. What is the capacity of the
tank in cubic feet ? In gallons ?
41. Separate 45 into two such parts that one part divided by
the other will give a quotient of 2 and a remainder of 9.
42. Separate 45 into two such parts that one part divided
by the other will give a quotient oi 5 axui %* wc&aEo&Kt. ^VSV.
162 MATHEMATICS
43. Separate 45 into two such parts that one part divided by
the other will give a quotient of 5 and a remainder of 48.
44. Two men, 58 miles apart, start at the same time to travel
toward each other. The first travels 7 miles in 2 hours, the
second travels 15 miles in 4 hours. How far does the first one
travel before they meet ?
45. The denominator of a fraction is 5 less than the numer-
ator. If 5 is added to the numerator the value of the fraction
is then \. Find the fraction.
46. Two men, A and B, 58 miles apart, start at the same
time to travel toward each other. A travels 7 miles in 2 hours
and B travels 15 miles in 4 hours. A meets with an accident
and is delayed 20 minutes. How far does B travel before
they meet ?
47. Separate a into two parts such that one part divided by
the other will give a quotient of g and a remainder of c.
48. Separate 42 into three parts such that the second shall
be four times the first and the third 4 times the second.
49. Separate c into three parts such that the second shall be
m times the first and the third m times the second.
50. A has $ 1200 out at interest, part at 6 % and the balance
at 5 °/ • The part at 6 % brings as much interest in 4 years as
the part at 5 % brings in 6 years. What is his total amount
of interest per year ?
., 4 (58 -a) 2x
15 7
Multiplying by 105,
28- 68 -28s = 30s.
58 x = 28 • 58.
35 = 28.
(Indicate such products and save computation.)
„ 24 — x Q , 5x M 5x — 2 . 7oj — 6 15aj-5
52. = y-i . 53. \ =. •
16 ^48 <> T 5x^ A3>
FRACTIONS 163
Supplemental Applied Mathematics
1. The specific gravity of ice (ratio of the weight of a cubic
foot of ice to that of a cubic foot of water) is .92. How much
water is there in a cake of ice 3' X 10.5" x 10.5" ? What will
the ice weigh ?
2. What is the third dimension of a cake of ice 1' X 1',
weighing 50 pounds ?
3. A piece of iron 4" x 8" x 1' is placed in a tank of water.
How many liters of water did it displace ?
4. One cubic foot of steel immersed in water weighs how
much ?
5. The specific gravity of sea water is 1.025, and that of ice
is .92. Find the difference in weight between one cubic foot of
sea water and one cubic foot of ice.
6. A piece of ice 3' X 1' X 1' is dropped overboard from an
ocean liner. How much of the ice is submerged in the ocean ?
7. A tank of water 18' x 8' and 6' deep is frozen to a depth
of 7". Find the value of the ice at 42 ^ per hundred pounds ?
a Air is 14.43 times as heavy as hydrogen gas. 8500 cubic
feet of hydrogen have been pumped into a balloon. What
weight will it lift ?
9. In testing 100 pounds of steam coal there was found
8.3 pounds of ash, .932 pounds
sulphur. What was the per cent of
carbon ?
10. The span of a roof is 42'.
The pitch of AC is 30°. The
member* CE bisects the angle
ACB. Find the lengths of all the
members used.
* The pieces used in forming a root ttwaa axe caX\sA. u Tttwnaoraw^ vv ^c^a
irons, " or "angles. "
i
164
MATHEMATICS
11. Find the lengths of the
angles if the roof in example
11 is of the form of the truss
shown in the accompanying
figure. (EFG is equilateral.)
12. A 1£" rainfall on 20
acres of land is how many
barrels of water?
13. A school building has eight drinking fountains, each
flowing 1£ gallons per minute. These are supplied by a pump
whose cylinder is 4" in diameter and 10" long. How many
strokes per minute must the pump make to keep these fountains
running ?
14. The illumination given by any light is inversely as the
square of the distance — = 2
I 2 d?
A light 6 feet from a table is moved 3 feet from the table.
How much more light does the table now receive ?
15. A 16 candle power and a 4 candle power electric light
are placed on opposite sides of a screen. The 4 candle power
is 3' from the screen. At what distance must the 16 candle
power be placed that each side of the screen may receive the
same amount of light ?
16. A chandelier directly over a table contains four 16 candle
power carbon filament lights. These lights are 4'-6" from the
ceiling. A new chandelier fitted with four tungsten lights
3-6" from the ceiling is put in. These tungsten lights give 1\
times as much light as the old 16 candle power. Does the table
receive more or less light and how much ?
17. A 16 candle power electric light is 5' above a table
and does not give sufficient light. To remedy this defect a wire
is attached to the socket and brought down to a reading lamp
containing a tungsten burner which is \& M afocrc^ the table.
How much more light does the table TeeeVvet
FRACTIONS
165
la In cooling iron shrinks about £" per lineal foot after it is
cast. A casting must be 2-1" in length and l'S n square after
cooling. What were its dimensions when cast ?
19. A gas engine cylinder is to be 4" in diameter and 4"
long. What is its size before cooling ?
20. A gas engine cylinder is to be 3£" in diameter and
4" long. Find dimensions of the pattern from which it is
cast.
21. What size bolt would you use in a 0.344" hole ?
22. What size bolt would you use in a 1.491" hole ?
23. A man was standing behind a target during target
practice. Those doing the firing were 1 mile away and the
velocity of their projectiles was 1150' per second. Did the
projectile strike the target before the
sound reached there? What was the
difference between the time the sound of
firing and the projectile reached the target ?
24. Estimate the weight per lineal foot
of the Steel Tee Eail in the accompanying
figure. Disregard the round corners ; con-
sider them as angles.
25. If a plumber needs to change the direction of a pipe by
45°, he calls the hypotenuse AC of the triangle ABC equal to
BC + A BC. What is the error when BC = 30" ? Which is
easier to compute, the steam fitter's method or the correct
method ?
26. In estimating material for a bias ruffle a dressmaker
calls the length of the bias f the width of the goods. What is
the error when the goods is 27" wide ?
27. A plumber makes a 45° turn across a hallway 10' wide.
The hall is 46' long and is piped the entire length. What
length of pipe ia used ?
166
MATHEMATICS
28. In example 27, how many feet of pipe would be needed
if a 30° angle were used in making the cross over ?
29. In a triangle ABC, right-angled at B, Z A is 30°. In
such triangles many mechanics estimate that AC is 1^ times
AB. How nearly correct is this when AB is 10'? Is this
method as easy to compute as the correct one, namely,
Let 2x=AC, then x = BC, and AB = V(2 xf - (x) 2 .
Let AB = a, then V(2 x) 2 — x 2 = a,
30. A house is 26 feet wide. The pitch of the roof is 30°.
Find the length of the rafters, no allowance being made for
extensions at the eaves.
31. The width of a house is 28' and the roof has a pitch of
30°. What length rafters must the carpenter cut if the rafters
project 14" beyond the house at the eaves?
32. Use dressmaker's rule in finding the number of yards of
goods necessary to make a 6-inch bias ruffle for a skirt 4 yards
around. The goods is 27"
wide and no strip of ruffling
is to be less than 35" inlength.
Let AFbe the piece of goods,
then AK= 36", the length of
each strip of ruffling. It will
therefore require 6 strips of
ruffling for this 4-yard skirt. The cut AB along the selvage
is 8". The amount of goods required is therefore 6x8" plus
the 27" (width of goods MO) wasted at the corner, or 2\ yards.
33. Find the cost of two 3" bias ruffles made from goods
30" wide and costing 95^ per yard, the skirt being 4 yards
around.
34. Silk may be purchased " on ttte \>\a& " TV^ avoids pay-
ing for the waste at the corner. ¥md t\ifc cosfc oi \to^\^ v \ss»&
A 8"B
FRACTIONS 167
ruffles to be used on a 6-yard silk skirt, silk 21" wide and
costing $ 1.10 per yard.
35. The braces for a billboard are to have an angle of 60°
with the ground. What length must the braces be cut if the
foot of the brace is 8' from the board ?
36. A piece of steel is 1£" by 5". It weighs 225 pounds.
Find its length.
37. The volume of a cube is .512 cubic foot. Find its surface.
38. The water in an irrigating ditch flows 3£ miles per
hour. It must supply a 160-acre farm with 1" of water per
week. What is the area of a cross section of the ditch ?
39. Water is flc!wing through an 8" water main at the rate of
10' per second. How many barrels of water will it deliver per
hour?
40. A farmer had a pond of 6 acres which was frozen to a
depth of 10". He sold the ice to a dealer at 12 ff per hundred
pounds. How much did he receive ?
41. One diagonal of a rhombus and a side of the rhombus are
each 18". Find the other diagonal, the angles, and the area.
42. A bar of copper 4" x 6" x 3' is drawn into a £" wire.
How long is the wire ? What does the coil weigh ?
43. The area of a cold-air box is to be £ less than the com-
bined areas of the hot-air pipes. One hot-air pipe is 10" in
diameter, one is 8", and the remaining six are each 6". Find
the area of the cold air box.
44. From a town C one train goes north at 32 miles an hour.
One hour later a second train goes east at 30 miles an hour.
How far apart are they 3 hours after the first train started ?
45. From a town C a train goes north at 32 miles an hour.
5 hours later a second train follows the first at 40 miles an
hour. How far apart are they 8 hours after the second train
started?
168 MATHEMATICS
46. Chocolate contains 12.9 % protein ; cocoa, 21.6 %. How
much chocolate will furnish as much protein as \ pound cocoa *
47. Chocolate contains 48.7% fat, and cocoa, 28.9%. One
half pound cocoa measures 2 cups. How many cups cocoa
will furnish as much fat as \ pound chocolate ?
48. As to the quantity of fat, which is cheaper to use, cocoa
or chocolate, if 8 ounces chocolate cost 22^ and \ pound cocoa
costs 25^?
49. If halibut costs 18^ per pound and 17.7% is refuse,
find cost per pound of edible portion.
50. Haddock costs 12^ per pound ; 51 % is refuse. Which is
cheaper fish, halibut or haddock ?
51. Boned and dried codfish sells for 16 f per pound ; dried
codfish for 10^ per pound. From the latter there is a loss of
20%. Which is cheaper?
52. Whitefish contains 12.8 % protein ; 43.6 % is refuse. It
sells for 16^ per pound. Porterhouse steak contains 19.1%
protein ; 12.7 % is refuse. It sells for 30^ per pound. Which
food contains more protein for less money ?
53. Bass contains 9.3 % protein ; 54.8 % is refuse. It sells
for 12 ^ per pound. Which kind of fish is cheaper, whitefish
or bass ?
54. Herring contains 11.2% protein; 42.6% is refuse; it
sell for 10^ per pound. Perch contains 7.3 % protein; 62.5%
is waste; it sells for 10^ per pound. Which fish contains
more protein for less money ?
55. Pike contains 7.9 % protein ; 57.3 % is refuse ; it sells
for 12 ^ per pound. Round steak contains 27.6 f protein ; it
sells for 16^ per pound. Which is the cheaper food ?
56. Canned salmon sells for 18 ^ per can ; it weighs 1\ pounds ;
contains 19.5% protein, 14.2% waste. Sardines sell for 25^
per can, which weighs llf ounces ; they contain 23.7 % protein
and 5% waste. Which is cheaper to ua^
FRACTIONS 169
57. Dried beef contains 39.2% protein. It costs 30^ per
pound. Which is cheaper, dried beef or salmon ?
58. Express graphically the edible quantities of haddock,
halibut, whitefish, bass, herring, perch, pike, and canned salmon
that can be purchased for 25^.
59. Express graphically the quantities of protein per pound
in these fish.
CHAPTER X
Proportion
174. Review § 173 and exercise 60. The relation of one
number to another is often expressed in fractional form.
These fractions are known as ratios. Thus, the ratio of 2 to 3
is written •}. This was first written 2-*- 3, then the division
sign was modified to 2:3, now the fractional form is con-
sidered best.
A proportion is the equality of two ratios. It is therefore
simply a fractional equation of two terms.
Thus, - = - is a proportion. This is read a divided by b is
equal to c divided by d.
The numerators are the antecedents.
The denominators are the consequents.
The first antecedent and the last consequent are the extremes.
The first consequent and the second antecedent are the means.
Thus, a and d are extremes and b and c the means.
Properties of Proportion
1. The product of the extremes is equal to the product of the
means.
This is easily proved by clearing the equation (proportion) ,
a_c
of fractions.
Whence, ad = be.
170
PROPORTION 171
2. If the product of two numbers is equal to the product of two
other numbers, a proportion may be formed making one product
the means and the other product the extremes.
Thus, xy = mc.
Dividing by y • to, ' — = - •
to y
Had you divided by y • c, the proportion would have been
* = ™. (See 6, §174.)
c y
3. If four quantities are in proportion, they are in proportion
by composition.
Let
a _c
b~ d
or
Then, ? + i = 5 + i f
b d
a 4- b c + d
b d
4. If four quantities are in proportion, they are in proportion
by division.
Let *? = *.
b d
Then, « _ 1 = £ _ 1
b d '
or a-b = c--d t
b d
5. If four quantities are in proportion, they are in proportion
by composition and division.
Let ? = £ . (1)
b d w
By 3, a±b = cj L d, (2)
b d
*a„ A a — b c — d /on
By 4, __^ = — (3)
b a
Dividing (2) by (3), «-±l = c i±A <4\
a— b c — d
172 MATHEMATICS
6. If four quantities are in proportion, they are in proportion
by alternation.
Thus, if ^ = -^ then 2 = ^.
b d c d
175. In a mean proportion the means are equal.
ax
Thus, - = - is a mean proportion.
x d
Solving x = Vad.
That is, a mean proportional between a and d is the square root of their
product.
The last consequent of a mean proportion is a third proportional to the
other two numbers.
Thus in - = -, d is a third proportional to a and x.
x d
The fourth proportional is the last consequent in such a pro-
portion as - = - . where no two terms are alike.
F b d
EXERCISE 61
1. Find a mean proportional between 4 and 9.
2. Find a mean proportional between 16 and 25.
3. Find a mean proportional between 289 and 256.
289 = x
x 256'
or, x = V289 . 266 • But Vab = Va Vb
Then, x = V289 . v'266
= 17 • 16 = 272.
4. Find the mean proportional between 121 and 729.
5. Two lines are 196' and 25', respectively. Find a line
equal to their mean proportional.
6. What is the mean proportional between 1.44" and 2.56".
7. Find a, fourth proportional to *2' , 2> y , &' .
PROPORTION 173
a Find a fourth proportional to 5, 9, 15.
9. Find a fourth proportional to 6, 9, 3.
10. Find a fourth proportional to 5, 7£, 9.
2 6
Note that in such proportions as - = - , x may be obtained by inspec-
3 x
tion. For since 6=3*2 (the first numerator), x must be 3 • 3 (the first
denominator). Or, reading vertically instead of horizontally, since the
first consequent is 1£ times the first antecedent, the second consequent
must be 1J times the second antecedent.
11. Find a third proportional to 2, 8.
12. Find a third proportional to 5, 15.
13. Find a third proportional to 5, 9.
14. Find a mean proportional between .0529 and 529.
15. Find a mean proportional between ' — — ^— and
x + 2
16. Find a mean proportional between — -^ and
x + 7 a^ + Sx-7
2ar s + 3a;-5
«
17. Find a fourth proportional to x 2 -f- 9 x -f 20, a* 2 — 3 x — 28,
and 2 x 2 + 19 x + 45.
18. In a semicircle, if a perpendicular is dropped to the
diameter, from a point in the circumference, the perpendicular
is a mean proportional between the segments of the diameter.
The diameter of the circle is 20 and the perpendicular to it
from the circumference is 8. Find the segments of the diameter.
19. The segments AB and BG of a diameter AC are 4" and
9" respectively, find the length of the perpendicular to AG
erected at B and extending to the circumference.
^ 2«+_8 = 5x + U m Solv6j using § 174 ^ g bef Qre cleari
2x — 3 5x — 11
of fractious.
174 MATHEMATICS
2i («+7)+<3y-i) aa I.
• (x + 7)-(3y-l) 2
5x—7y= — 9.
(Use § 174, 5, on the first equation before clearing it of fractions.)
Ratio plays a very important part in science, though the ratio idea is
often disguised to such an extent by the scientific notation that the pupil
thinks in other terms than those of ratio or measurement.
For example, the mysteries of Specific Gravity disappear when one
feels that the specific gravity is simply the ratio of a volume of some sub-
stance to an equal volume of some substance taken as a standard.
The standard for liquids and solids is water.
One cubic centimeter (c.c.) of water weighs 1 gram, or 1 cu. ft. weighs
62J pounds.
For gases the standard is usually hydrogen ; sometimes air, which is
14.44 times as heavy as hydrogen, is used.
Ex. A cubic foot of steel weighs 490 lb. Find specific gravity of steel.
Specific gravity of steel = ±^ = 7.84.
6 J ■ 62.5
It is customary to write specific gravity in a decimal form, not as a
common fraction.
Units to be remembered :
1" • = 2.54 centimeters.
1 liter = 1000 cubic centimeters (c.c).
1 kilogram = 1000 grams.
1 c.c. water weighs 1 gram.
1 liter hydrogen weighs 0.09 gram.
Specific gravity air (hydrogen standard) is 14.44.
22. Ice weighs 57.5 pounds to the cubic foot. Find its spe-
cific gravity.
23. The specific gravity of oak is 0.8. Find' the weight of
1 cubic foot.
24. A cubic foot of lead weighs 706 pounds. Find its specific
gravity.
25. A cubic foot of copper weighs 550 pounds. Find its
specific gravity.
26. The specific gravity of aluminum is 2.6. Steel is how
many times as heavy ?
LIST OF CONSTRUCTIONS
Pagb
63. To draw a straight line equal to a given straight line ... 49
67. To construct an angle equal to a given angle ...... 50
72. To bisect a given angle 53
80. To construct a triangle when three sides are given .... 56
92. To draw a perpendicular bisector of a straight line . ... 64
93. To draw a perpendicular to a line from any point in the line . 65
96. To draw a perpendicular to a line from a given point with-
out the line ' 67
119. To draw a straight line parallel to a given straight line . . 97
m
LIST OF THEOBEMS
Lines
Theorem VII Paqb
91. If two straight lines intersect, the vertical angles are equac - 64
Theorem VIII
94. If a perpendicular is erected at the middle point of a line,
I. Any point in the perpendicular is equidistant from the extremi-
ties of the line
II. Any point not in the perpendicular is unequally distant f ran
the extremities of the line 65
Theorem IX
97. From a point without a line but one perpendicular can be
drawn to the line 67
Theorem XII
100. If two unequal oblique lines drawn from a point in a per-
pendicular to the line, cut off unequal distances from the foot of the
perpendicular, the more remote is the greater 70
Theorem XIII
101. If oblique lines are drawn from a point to a straight line and
a perpendicular is drawn from the point to the line,
I. Two equal oblique lines cut off equal distances from the foot
of the perpendicular
II. Hie greater of two unequal oblique lines cuts off the greater
distance from the foot of the perpendicular . 71
Theorem XIV
120. Two lines parallel to the same line are parallel to each other 97
Theorem XV
121. A line perpendicular to one of two parallels is perpendicu-
lar to the other 98
176
LIST OF THEOREMS 177
Theorem XVI _
Fags
124. If two parallel lines are cut by a transversal, the alternate-
interior angles are equal 100
Theorem XVII
125. If two lines are cut by a transversal, and the alternate-
interior angles are equal, the lines are parallel 101
Theorem XXI
131. Any point in the bisector of an angle is equidistant from
the sides of the angle 106
Theorem XXVI
137. If a series of parallels intercept equal parts on one transver-
sal, they intercept equal parts on every transversal Ill
Theorem XXVII
138. The line joining the middle points of two sides of a triangle
is parallel to the third side and equal to one half of it 112
Theorem XXVIII
139. The line joining the middle points of the non-parallel sides
of a trapezoid is parallel to the bases and equal to one half of their
sum 113
Lines which Meet in a Point
Theorem XXIX
140. The bisectors of two of the angles of a triangle intersect on
the bisector of the third angle 113
Theorem XXX
142. The perpendiculars erected at the middle points of the sides
of a triangle meet in a point which lies in the perpendicular bisector
of the third side 115
Theorem XXXI
144. The three altitudes of a triangle meet in a common point . 116
Theorem XXXII
145. Two medians of a triangle meet in a point of the third
median ^^^
178 MATHEMATICS
Triangles
Theorem I Paqib
70. Two triangles are equal when two sides and the included angle
of one are equal respectively to two sides and the included angle of
the other 51
Theorem II
71. Two triangles are equal when a side and two adjacent angles
of the one are equal respectively to a side and two adjacent angles of
the other 52
Theorem III
78. In an isosceles triangle the angles opposite the equal sides are
equal 55
Theorem IV
81. Two triangles are equal when three sides of one are equal re-
spectively to three sides of the other 57
Theorem V
89. Any side of a triangle is greater than the difference of the
other two sides . 62
Theorem VI
90. The sum of two sides of a triangle is greater than the sum of
two lines drawn from any point within the triangle to the extremities
of the third side of the triangle 63
Theorem X
98. Two right triangles are equal when the hypotenuse and an
acute angle of the one are equal, respectively, to the hypotenuse and
acute angle of the other 68
Theorem XI
99. Two right triangles are equal when the hypotenuse and leg of
the one are equal, respectively, to the hypotenuse and leg of the other 69
Theorem XVIII
126. TAe sum of the angles of a triangle is equal to Iijdo •n.cjh.t
an^/es "\B&
LIST OF THEOREMS 179
Theorem XIX
Page
128. If two angles of a triangle are equal, the triangle is isosceles 105
Theorem XX
129. If two sides of a triangle are unequal, the angles opposite
are unequal, and the greater angle lies opposite the greater side . 105
Theorem XXII
132. If two triangles have two sides of one equal respectively to
two sides of the other, and the included angle of the first greater than
the included angle of the other, the third side of the first is greater
than the third side of the second 108
Parallelograms
Theorem XXIII
134. In a parallelogram the opposite sides are equal, and the
opposite angles are equal 110
Theorem XXIV
135. Two parallelograms are equal if two sides and the included
angle of one are respectively equal to two sides and the included
angle of the other Ill
Theorem XXV
136. If the opposite sides of a quadrilateral are equal, the figure
is a parallelogram Ill
Polygons
Theorem XXXIII
149. The sum of the angles of any polygon is equal to twice as
many right angles as the polygon has sides, less four right angles . 117
Theorem XXXIV
150. The sum of the exterior angles of a polygon formed by pro-
ducing one side at each vertex of a polygon is equai to four riafct
angles . ••%%••**» V^
INDEX
Abcissa, 93.
absolute term, 95. *
absolute value, 17.
acute angle, 53, 60.
acute-angled triangle, 55.
addition, 16.
of fractions, 148.
adjacent angles, 49.
algebraic expression, integral, 152.
alternate exterior angles, 99.
alternate interior angles, 99.
altitude of triangle, 57, 98.
angle, 49.
acute, 53, 60.
alternate-exterior, 99.
alternate-interior, 99.
bisector, 53.
complementary, 61.
exterior, 98, 99.
exterior-interior, 99.
interior, 99.
obtuse, 53, 60.
of polygon, 116.
right, 53.
supplementary, 61.
supplementary adjacent, 61.
vertical, 61.
antecedent, 170.
applied mathematics, 27, 44, 72,
119, 139, 163.
arc, 48.
axioms, 11, 50.
Base of triangle, 57.
binomial, 2.
square of, 124.
bisector of angle, 53.
- of line, 64.
brace, 7.
bracket, 7.
88,
Circle, 48.
circumference, 48.
coefficient, 3.
common factor, 6, 127.
common multiple, 147.
complementary angles, 61.
complex fractions, 156.
conditional equation, 10.
consequents, 170.
constants, 95.
coordinates, 93.
curved line, 48.
Decimals, 27.
degree, 94.
diagonal, 109.
difference, 19.
of squares, 122.
digit, prime, 1.
dividend, 23.
division, 23.
by zero, 158.
fractions, 155.
polynomials, 81.
divisor, 23.
Elimination, 35.
equation, 10.
condition, 10.
equivalent, 96.
fractional, 157.
geometric, 11.
identical, 11.
inconsistent, 95.
root of, 10.
simultaneous, 35, 94.
solution of, 10.
equiangular polygon, 117.
I equilateral tnasi$&,TO.
180
INDEX
181
equivalent equations, 96,
exponent, 24.
exterior angle, 98.
extremes, 170.
Factor, 2, 122.
factoring, hints on, 134.
factors, common, 6, 127.
prime, 5.
fractional equations, 157.
fractions, 143.
addition, 148.
complex, 156.
decimal, 27.
division, 155.
improper, 151.
multiplication, 153.
reduction, 27, 143.
Geometric equation, 11.
solid, 47.
surface, 47.
geometry, plane, 48.
graphs, 92.
Hints on factoring, 134.
housekeeping problems, 32, 45, 72, 88,
120, 141, 166.
Identical equation, 11.
improper fractions, 151.
inconsistent equations, 95.
independent term, 95.
indirect method, 71.
inequalities, 33.
solution of, 42.
integral expressions, 152.
interior angles, 99.
isosceles* trapezoid, 109.
isosceles triangle, 55.
Line, 47.
broken, 48.
curved, 48.
parallel, 95.
sect, 47.
segment, 47.
locus of points, 106.
lowest common multiple, 147.
Mathematics applied, 27, 44, 72, 88,
119, 139, 163.
means, 170.
method, indirect, 71.
minuend, 19.
minutes, 61.
mixed numbers, 152.
multiple, common, 147.
multiplication, 22, 76.
fractions, 153.
polynomials, 76.
sign rule, 23.
Negative numbers, 15.
number, mixed, 152.
negative, 15.
positive, 15.
prime, 1.
system, 1.
Oblique angle, 53.
oblique angle triangle, 55.
obtuse angle, 53, 60.
obtuse angle triangle, 55.
oral review, 8.
ordinate, 94.
origin, 94.
Parallel lines, 95.
parallelogram, 109.
parenthesis, 7.
perpendicular, 53.
perpendicular bisector, 64.
plane, 48.
polygon, 116.
mutually equilateral, 117.
regular, 117.
polynomial, 2.
division, 81.
multiplication, 77.
positive numbers, 15.
prime digit, 1.
factor, 5.
number, 1.
principles of fractions, 143.
problems, housekeeping, 32, 45, 72, 88,
120, 141, 166.
182
INDEX
Quadrilateral, 109.
quadrinomial, 2.
quotient, 23.
Radius, 48.
reading of figures, 109.
rectangle, 109.
reduction of fractions, 27, 145.
regular polygon, 117.
review, oral, 8.
rhomboid, 109.
rhombus, 109.
right angle, 53.
right triangle, 55.
root of equation, 11.
rule, sign, 17, 20, 23.
subtraction, 20.
Scalene triangles, 55.
seconds, 61.
sect of line, 47.
segment of line, 47.
shop problems, 27, 44, 72, 89, 119, 139,
163.
sign rule, addition, 17.
division, 23.
multiplication, 23.
subtraction, 20.
similar terms, 2.
simultaneous equations, 35, 94.
solid, geometric, 47.
solution, equations, 11.
factoring, 137.
inequalities, 42.
square, 109.
of binomial, 124.
substitution, 37.
subtraction, 18.
sule, 20.
subtrahend, 19.
supplementary angles, 61.
surface, geometric, 47.
straight line, 47.
Term, 2.
degree of, 95.
terms, similar, 2.
theorem, 50.
transposition, 34.
transversal, 99.
trapezium, 109.
trapezoid, 109.
triangle, 50.
acute, 55.
altitude, 57.
base, 57.
equilateral, 55.
isosceles, 55.
oblique, 55.
obtuse, 55.
right, 55.
scalene, 55.
vertical angle, 98.
vertices, 50, 98.
trinomial, 2.
trinomial square, 124.
Value, absolute, 17.
variables, 35, 95.
vertical angles, 61.
of triangle, 98.
vertices of triangle, 50.
vertex, 49.
of triangle, 98.
vinculum, 7.
Zero, 1.
zero division, 158.
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